MODULE : 1
Lecture 1 :
Key words :
Scalar, Vector, Field, position vector, dot product, cross product, Right hand rule
Multiple Choice Questions :
1. A scalar quantity
a. is specified only by its magnitude
b. does not have unit
c. always has a mass
d. is not dimensioned.
2. Which of the following is a scalar?
a. velocity
b. current density
c. electromotive force
d. electric field
⃗⃗| = |𝐴⃗ − 𝐵
⃗⃗|, then the angle between the vectors 𝐴⃗ and 𝐵
⃗⃗ is
3. If |𝐴⃗ + 𝐵
a.
b.
c.
d.
zero
450
600
900
⃗⃗ having magnitudes a and b are added, the magnitude of the
4. When two vectors 𝐴⃗ and 𝐵
resultant vector
a. is equal to a+b
b. is equal to a-b
c. cannot be greater than √𝑎2 + 𝑏 2
d. cannot be greater than a+b
⃗⃗ are such that the magnitude of their dot product is equal to magnitude of
5. If two vectors 𝐴⃗ and 𝐵
their cross product, the angle between the vectors is
a. zero
b. 450
c. 600
d. 900
⃗⃗. If none of the vectors is a null
6. A vector 𝐶⃗does not lie in the plane containing vectors𝐴⃗ and 𝐵
⃗⃗ + 𝐶⃗
vector, it follows that the sum of the three vectors 𝐴⃗ + 𝐵
a. is not a null vector
b. can be a null vector
⃗⃗
c. may lie in the plane containing the vectors𝐴⃗ and 𝐵
⃗⃗ and either of the vectors 𝐴⃗ and 𝐵
⃗⃗.
d. may lie in the plane containing the vector 𝐴⃗ + 𝐵
Problems :
Determine the angle between the vector 𝑖̂ + 𝑗̂ + 𝑘̂ and the unit vector 𝑖̂.
Find a unit vector perpendicular to the vector 4𝑖̂ − 3𝑗̂.
̂.
Determine a vector having a magnitude 1/2 which is anti - parallel to the vector 10𝑖̂ − 5𝑗̂ + 10𝑘
For a non-zero vector 𝑎⃗ = 𝑎1 𝑖̂ + 𝑎2 𝑗̂ + 𝑎3 𝑘̂, the cosine of the angle between the vector 𝑎⃗ and
the unit vectors 𝑖̂, 𝑗̂ and 𝑘̂ are known as the “direction cosines” of the vector 𝑎⃗ :
𝑎⃗ ∙ 𝑖̂
𝑎⃗ ∙ 𝑗̂
𝑎⃗ ∙ 𝑘̂
𝐜𝐨𝐬 𝜶 =
, 𝐜𝐨𝐬 𝜷 =
, 𝐜𝐨𝐬 𝜸 =
|𝒂|
|𝒂|
|𝒂|
2
2
Show that the direction cosines satisfy the relation cos 𝛼 + cos 𝛽 + cos2 𝛾 = 1 and find the
direction cosine of the vector 3𝑖̂ + 4𝑗̂ + 5𝑘̂.
5. Find the angle between the body diagonal AC of the cube with its face diagonal AB.
1.
2.
3.
4.
C
A
B
6. Find the area of a triangle whose onevertex is at the origin and the position vectors of the
remaining vertices are 𝑖̂ + 4𝑗̂ and 2𝑖̂ + 2𝑗̂ + 2𝑘̂.
7. Three edges of a parallelepiped which meet at the origin are defined by position vectors 𝑖̂ + 4𝑗̂
,2𝑖̂ + 2𝑗̂ + 2𝑘̂ and −𝑖̂ + 4𝑗̂.
Answers to Multiple Choice Questions
1. a
2. c
3. d
4. d
5. b
6. a
Answers to Problems
1
1. cos−1( 3)
√
2.
3.
4.
3
4
𝑖̂ + 5 𝑗̂
5
1
1
1
− 3 𝑖̂ + 6 𝑗̂ − 3 𝑘̂
3
cos 𝛼 = 5 2 , cos 𝛽
√
=5
5
√2
, cos 𝛾 = 5
5
√2
2
5. cos−1( 6)
√
6. 2√2
7. 10
Hints for solutions to problems :
1. Find the dot product of the vectors and divide it by the product of magnitude of the vectors.
2. For a vector to be perpendicular to another vector, their dot product should vanish. Take the
vector to be 𝑖̂ + 𝑏𝑗̂ , equate the dot product to zero. This will give you a relation between a
andb. Since the vector is to be a unit vector 𝑎2 + 𝑏 2 =1.
̂ so as to be anti-parallel to the
3. The required vector has to be proportional to −10𝑖̂ + 5𝑗̂ − 10𝑘
given vector. This vector has a magnitude 15. Thus to get magnitude of 1/2 , divide this by 15/2.
4. Direct use of formula.
⃗⃗⃗⃗⃗⃗ and 𝐴𝐵
⃗⃗⃗⃗⃗⃗ . Determine their dot product and find the angle as in Problem 1.
5. Find the vectors 𝐴𝐶
6. Area vector is cross product of the vector s representing the edges. Its magnitude represents the
area.
7. Volume is given by taking the dot product of the vector of the third edge with the area vector.
Errors to be corrected :
1. Page 7 line 2 : inside the bracket (math it should be aligned) should read 0 ≤ 𝜃 ≤ 𝜋
2. Page 10 : line 3 : raise the line “ points in the direction …. Called the” to align with AXB
3. Last line before the figure –dS should be aligned with the word “element”
4. Example 1 : (Page 14 pop up) all underlines should be removed.
Lecture 2 :
Key Words : Cartesian coordinates, spherical coordinates, cylindrical coordinates, Jacobian, coordinate
transformation.
Multiple choice questions :
1. Rectangular coordinate system is also known as
a. Space coordinate system
b. Polar coordinate system
c. Cartesian coordinate system
d. Planar coordinate system
2. The least distance between two points on the equatorial circle on the surface of the earth
having polar coordinates (R,0) and (R, π/2) with respect to the origin at the centre of the Earth
is
3.
4.
5.
6.
7.
8.
a. R
b. R/2
c. Rπ/2
d. Rπ
The range of azimuthal angle φ in the spherical polar coordinates is
a. [0,2π]
b. [0,π]
c. [0,π/2]
d. [-π, +π]
The equation to a surface in spherical coordinates is given by φ=π/3. The surface is
a. A sector of a circle
b. A cone making an angle of π/3 with the z-axis
c. A vertical plane making an angle of π/3 with the z-axis
d. A vertical plane making an angle of π/3 with the x-axis
The equation to a surface in spherical coordinates is given by θ=π/3. The surface is
a. A sector of a circle
b. A cone making an angle of π/3 with the z-axis
c. A vertical plane making an angle of π/3 with the z-axis
d. A vertical plane making an angle of π/3 with the x-axis
In two dimensions, for the transformation from a pair of variables (x,y) to a new pair (u,v), the
Jacobian is
a. The ratio of the elemental area dudv to the area dxdy
b. The ratio of the elemental area dxdy to the area dudv
c. Only depends on the area element dxdy
d. Only depends on the new area element dudv
The Jacobian for transformation from three dimensional Cartesian coordinates to spherical polar
coordinates is
a. 𝑟 sin 𝜃
b. 𝑟 2 sin 𝜃
c. 𝑟 2 sin 𝜃 cos 𝜑
d. 𝑟 cos 𝜃
Expressed in spherical coordinates, the equation 𝑥 2 + 𝑦 2 + 𝑧 2 = 4𝑧 becomes
a. 𝑟 = 4 cos 𝜃 sin 𝜑
b. 𝑟 = 4 sin 𝜃 cos 𝜑
c. 𝑟 = 4cos 𝜃
d. 𝑟 = 4 sin 𝜃
Problems
1
√1−𝑥 2
1. Rewrite ∫−1 𝑑𝑥 ∫0
√𝑥 2 + 𝑦 2 𝑑𝑦 in polar coordinates.
2. Convert (𝑥, 𝑦, 𝑧) = (1, −1, √2) to (a) spherical coordinates and (b) cylindrical coordinates.
𝜋
3. Sketch the region 0 ≤ 𝜃 ≤ 2 , 𝜌 ≤ 1 of the cylindrical coordinate system. Define the region
in spherical polar coordinates.
𝜋 𝜋
4 3
4. Convert (4, , ) in spherical coordinates to (a) rectangular coordinates and (b) cylindrical
coordinates.
𝑣
5. Find the Jacobian of the transformation from the xy plane to the uvplane :𝑥 = 𝑢2 , 𝑦 =
𝑣2
.
𝑢
6. Evaluate the integral ∬|𝑥| |𝑦|𝑑𝑥𝑑𝑦 over a circular region of radius R centered at the origin.
7. Using an appropriate change of variables calculate the double integral ∬(𝑥 + 2𝑦)2 𝑑𝑥𝑑𝑦
1
over the region defined by the parallelogram with vertices at (0,0), (1,1/2), (1, − 2) , (2,0).
y
1/2
2
0
x
8. Using cylindrical coordinates, calculate the volume of a hemisphere.
Answers to Multiple choice questions :
1. (c) 2. (c) 3. (a) 4. (d) 5. (b) 6. (b) 7. (b) 8. (c)
Hints for solutions to problems :
1. The upper limit on y integration satisfies 𝑥 2 + 𝑦 2 = 1. Since 𝑦 ≥ 0, the region of integration is a
semicircle of radius 1 with −1 ≤ 𝑥 ≤ +1 as its diameter. The area element dxdy becomes
1
𝜋
𝑟𝑑𝑟𝑑𝜃 in polar. Thus, the integral becomes ∫0 𝑑𝑟 ∫0 𝑟 2 𝑑𝜃.
𝜋 7𝜋
)
4
2. Answer :(a) (2, 4 ,
, (B) (√2,
7𝜋
, √2).
4
One has to be careful in fixing the value of φ in
spherical polar (or of θ in cylindrical). Since the point is in the fourth quadrant of the xy plane,
tan 𝜑 = −1 has the solution
7𝜋
.
4
3. In xy plane it is quarter of a disk 𝑥 2 + 𝑦 2 ≤ 1, z unbound. Thus the region defines quarter of an
infinite circular cylinder. In spherical polar 𝑥 2 + 𝑦 2 ≤ 1 becomes 𝑟 2 sin 𝜑 ≤ 1.
𝜋
3
4. Answer : (𝑥, 𝑦, 𝑧) = (√2, √6, 2√2); (𝑥, 𝑦, 𝑧) = (√2, √6, 2√2); (𝜌, 𝜃, 𝑧) = (2√2, , 2√2)
5. The Jacobian is given by the inverse of the determinant
𝑢4
𝜕𝑥
𝜕𝑢
|𝜕𝑦
𝜕𝑢
𝜕𝑥
𝜕𝑣
|
𝜕𝑦
𝜕𝑣
𝑣2
= −3 𝑢4. Thus the required
1
Jacobian is − 3𝑣 2 ≡ − 3𝑥 2.
6. The required integral is equal to 4 times the integral over the first quadrant, where both x and y
are positive. Convert the integral to polar. Value of the integral is 𝑅 2 /2.
7. Let 𝑢 = 𝑥 + 2𝑦, 𝑣 = 𝑥 − 2𝑦. The Jacobian is 1/4 . (Remember that inside integrals only the
1
2
2
3
magnitude of the Jacobian is to be taken). The required integral is ∫0 𝑢2 𝑑𝑢 ∫0 𝑑𝑣 = .
4
4
8. In cylindrical coordinates, the equation to the sphere is 𝜌2 + 𝑧 2 = 𝑅 2. Range of both ρ and z
vary from 0 to R (For a sphere, z would vary from –R to +R). Volume element in the cylindrical
coordinates is 𝜌𝑑𝜌𝑑𝜃𝑑𝑧. The integral over the hemisphere is
2𝜋
𝑅
√𝑅2 −𝑧 2
∫0 𝑑𝜃 ∫0 𝑑𝑧 ∫0
𝑅 𝑅2 −𝑧 2
𝜌𝑑𝜌 = 2𝜋 ∫0
2
𝑑𝑧 = 2𝜋
𝑅3
.
3
Lecture 3 :
Key words : line integral, work, parameterization, flux, surface integral
Problems :
1. Evaluate the integral ∫𝐶 𝑥𝑦 2 𝑑𝑥 where C is a quarter circle of radius 2 on the first quadrant
with the centre of the circle being at the origin.
2. Evaluate the work done by a force 𝐹(𝑥, 𝑦) = 𝑦𝑖̂ + 𝑥𝑗̂ acting along the curve 𝑦 = ln 𝑥 from
the point (1,0) to (𝑒 2 , 1) in a counterclockwise fashion.
3. Evaluate the integral∫𝐶(𝑥 2 + 𝑦 2 )𝑑𝑥 + 2𝑥𝑦𝑑𝑦 where C is a semicircle of radius 1 about the
origin (in a counterclockwise fashion).
4. Show that the integral ∫𝐶(𝑥𝑦 2 )𝑑𝑥 + (𝑥 2 𝑦 + 2𝑥)𝑑𝑦 around any square of side a depends
only on the area of the square.
5. Find the work done in moving an object in a force field given by 𝐹⃗ = 𝑦𝑖̂ + 𝑧 2 𝑗̂ + 𝑥 2 𝑘̂ along
a closed curve 𝑟⃗⃗⃗(𝑡) = sin 𝑡 𝑖̂ + cos 𝑡 𝑗̂ + 𝑡𝑘̂ ; 𝑡: [0,2𝜋].
6. Evaluate the line integral ∮ (𝑥 2 − 𝑦 2 )𝑑𝑙 where the contour is along 𝑥 = 3 cos 𝑡, 𝑦 = 3 sin 𝑡.
7. A vector field 𝐹⃗ representing the velocity field of a fluid is given by 𝐹⃗ = 𝑥 𝑖̂ + 2𝑧𝑗̂ + 𝑦𝑘̂.
Find the flux of the filed through the surface S given by the part of the plane 𝑥 + 𝑦 + 𝑧 = 1
in the first octant and oriented upward.
8. Let S be a circular cylinder of radius 3 and height 4. If 𝐹⃗ = 𝑥𝑖̂ + 2𝑦𝑗̂ + 𝑧 2 𝑘̂ , find the surface
integral of the force field through the curved surface of the cylinder.
9. Find the flux of 𝐹⃗ = −𝑦𝑖̂ + 𝑥 2 𝑗̂ + 3𝑧𝑘̂ out of the closed surface bounded by 𝑧 = 𝑥 2 + 𝑦 2
and z=1.
10. Evaluate the flux of the vector field 𝐹⃗ = 𝑦𝑖̂ − 𝑥𝑗̂ + 𝑧𝑘̂ through a cone 𝑧 =
√𝑥 2 + 𝑦 2 ;
0 ≤ 𝑧 ≤ 2.
Hints for solutions to problems :
1. Parameterize 𝑥 = 2 cos 𝑡; 𝑦 = 2 sin 𝑡 so that 𝑑𝑥 = −2 sin 𝑡 𝑑𝑡. Taking the contour in the
anticlockwise direction, t varies from π/2 to 0. The line integral, taken in the anticlockwise
fashion ,is+4.
𝑒2
1
2. ∫𝐶 𝐹⃗ ∙ 𝑑𝑙⃗ = ∫ 𝐹𝑥 𝑑𝑥 + ∫ 𝐹𝑦 𝑑𝑦 = ∫ 𝑦𝑑𝑥 + ∫ 𝑥𝑑𝑦 = ∫1 ln 𝑥 𝑑𝑥 + ∫0 𝑒 𝑦 𝑑𝑦 = 𝑒 2 + 𝑒
3. Parameterize 𝑥 = 2 cos 𝑡; 𝑦 = 2 sin 𝑡. Answer = 2/3
4. Without loss of generality, take the vertices of the square at (0,0), (a,0), (0,a) and (a,a). Calculate
each line integral. Along the line joining (0,)0 and (a,0) , y=0 and dy =0, the integral is zero. From
𝑎
𝑎
(a,0) to (a,a), x=constant=a, dx=0. The integral is ∫0 𝑎2 𝑦𝑑𝑦 + ∫0 2𝑎𝑑𝑦 =
𝑎4
2
+ 2𝑎2 . Add two
other contribution to show that the resulting integral is proportional to the area of the square.
𝑑𝑟⃗
𝑑𝑥
𝑑𝑦
𝑑𝑧
5. 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 𝑧 = 𝑡. ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫ 𝐹⃗ ∙ 𝑑𝑡 𝑑𝑡 = ∫ (𝐹𝑥 𝑑𝑡 + 𝐹𝑦 𝑑𝑡 + 𝐹𝑧 𝑑𝑡 ) 𝑑𝑡. The value of
the integral is 2𝜋 + 4𝜋 2 .
6. Answer : 0
7. The unit normal to the given plane is
∫𝐹⃗ ∙ 𝑑𝑆⃗ =
𝑆
1
√3
1
(𝑖̂ +
√3
𝑗̂ + 𝑘̂ ). Thus flux =
∫(𝑥 + 2𝑧 + 𝑦)𝑑𝑆 =
𝑆
1
√3
∫(2 − 𝑥 − 𝑦)𝑑𝑆
𝑆
z
Since the integrand is a function of x and y,
it is convenient to convert the integral to a
surface integral in the xy plane. This is
done by observing that if z=g(x,y) is the
equation to a surface S which projects on to
the region R of the x-y plane, we have,
𝐹⃗
y
x+y=
1
x
∫𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∫ 𝑓(𝑥, 𝑦, 𝑔(𝑥, 𝑦))√(
𝑆
𝑆
𝜕𝑔 2
𝜕𝑔 2
) + ( ) + 1 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
In this case z=g(x,y)=1-x-y. Thus
1
∫(𝐹⃗ ∙ 𝑑𝑆⃗ =
√3
𝑆
∫(2 − 𝑥 − 𝑦)𝑑𝑆 =
𝑆
1
√3
∫(2 − 𝑥 − 𝑦)√3 𝑑𝑥𝑑𝑦
𝑆
{1−𝑥}
1
(2 − 𝑥 − 𝑦)𝑑𝑦 =
= ∫ 𝑑𝑥 ∫
0
0
2
3
8. The outward normal to the curved surface of the cylinder is given by 3 cos 𝜃𝑖̂ + 3 sin 𝜃𝑗̂.
The surface integral over the closed surface is given by
4
2𝜋
∫ 𝑑𝑧 ∫ (3 cos 2 𝜃 + 6 sin2 𝜃)3𝑑𝜃 = 108𝜋
0
0
3 cos 𝜃𝑖̂ + 3 sin 𝜃𝑗̂
9. Consider the curved surface first. The unit vector which points upward (i.e. with a positive
component along the z-axis is given by 𝑛̂ =
1
∫ 𝐹⃗ ∙ 𝑛̂𝑑𝑆 = ∫
𝑆 √4𝑥 2
𝑆
̂
−2𝑥𝑖̂−2𝑦𝑗̂ +𝑘
. Thus
√4𝑥 2 +4𝑦 2 +1
+ 4𝑦 2 + 1
(−2𝑥𝑦 − 2𝑦𝑥 2 + 3𝑥 2 + 3𝑦 2 ) 𝑑𝑆
= ∬(−2𝑥𝑦 − 2𝑦𝑥 2 + 3𝑥 2 + 3𝑦 2 )𝑑𝑥𝑑𝑦
We can parameterize the integral by 𝑥 = 𝑟 cos 𝜃; 𝑦 = 𝑟 sin 𝜃 ; 0 ≤ 𝑟 ≤ 1, 0 ≤ 𝜃 ≤ 2𝜋. The
integral can be easily calculated to give 3𝜋/2.
The normal to the cap of the hemisphere is along the negative z direction. Since z=0 on this cap,
it does not contribute to the surface integral.
10. The problem of calculating surface integral over a cone is very similar except that the radius r
depends on the z-coordinate. Unit normal to the slanting surface is 𝑛̂ =
̂𝑧
−𝑖̂𝑥−𝑗̂ 𝑦+𝑘
.
√2𝑧
the method outlined earlier, we can parameterize (the slant angle is π/4
𝑥 = 𝑟 cos 𝜃; 𝑦 = 𝑟 sin 𝜃 ; 𝑧 = 𝑟; 0 ≤ 𝑟 ≤ 1, 0 ≤ 𝜃 ≤ 2𝜋
2
𝑥𝑦+𝑥𝑦+𝑧
2𝜋
Thus ∫𝑆 𝐹⃗ ∙ 𝑛̂𝑑𝑆 = ∬ (
) 𝑑𝑥𝑑𝑦 = ∬(2𝑟 cos 𝜃 sin 𝜃 + 𝑟)𝑟𝑑𝑟𝑑𝜃 =
𝑧
3
Following
z
𝑛̂
y
x
Lecture 4 :
Problems :
1. Use the fundamental theorem of calculus to compute the derivative of the integral 𝑓(𝑥) =
𝑥3
2
∫𝑥 𝑒 −𝑡 𝑑𝑡.
𝐵
⃗⃗⃗⃗⃗ ∙ ⃗⃗⃗⃗
2. Let 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥𝑦 2 . Find ∫𝐴 ∇𝑓
𝑑𝑙 , over the arc of a circle of radius 1 connecting A
(1,0) and B (0,1).
B
A
3. Find the work done by a conservative force field 𝐹⃗ = (4𝑥 3 𝑦 3 + 3)𝑖̂ + (3𝑥 4 𝑦 2 + 1)𝑗̂ on a
particle acting along the straight-line joining the point A to B as shown for problem 2.
4. Find the directional derivative of 𝑓(𝑥, 𝑦, 𝑧) = 2𝑥 2 𝑦 3 + 3𝑥𝑦 + 𝑧 2 along the direction 𝑖̂ + 𝑗̂ +
𝑘̂ at the point (1,1,1).
5. Find the unit vector normal to the surface 4𝑥 2 + 𝑦 2 + 9𝑧 2 = 18 at the point (1,1,1).
6. Calculate the gradient of 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑦 2 in Cartesian coordinate system as well as in the
spherical polar coordinates and show that they give the same results.
7. Verify divergence theorem for the vector field 𝐹⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + (𝑧 − 1)𝑘̂ over a hemisphere
𝑥 2 + 𝑦 2 + 𝑧 2 ≤ 𝑅 2, the bottom cap being at z=0.
8. Evaluate the surface integral of the vector field 𝐹⃗ = 3𝑥𝑧 2 𝑖̂ + 5𝑦 2 𝑗̂ − 𝑧 3 𝑘̂ over the region
𝑥2
,𝑥
2
bounded by the surfaces (see figure) 𝑧 = 𝑦, 𝑧 = 4 − 𝑦, 𝑧 = 2 −
= 0, 𝑧 = 0 .
z
z=y
x=0
y
z=4-y
x
𝑧 = 2 − 𝑥 2 /2
9. Find the surface integral of 𝐹⃗ = 𝑦 2 𝑖̂ + 2𝑥𝑧𝑗̂ + (𝑧 − 1)2 𝑘̂ over a region bounded by a
cylinder 𝑥 2 + 𝑦 2 = 4 and the planes z=1 and z=2.
10. Let S be an open cone 𝑧 = √𝑥 2 + 𝑦 2 with 𝑧 ≤ 4 . Calculate ∫ 𝐹⃗ ∙ 𝑑𝑆⃗ over the surface of the
𝑆
cone, where 𝐹⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂.
𝑘
11. Find the divergence of 𝐹⃗ = 𝑛 𝜌̂, and find n for which the divergence vanishes for all ρ>0.
𝜌
Hints for solutions to problems :
2
1. If F(x) is anti-derivative of 𝑒 −𝑥 , 𝑓(𝑥) = 𝐹(𝑥 3 ) − 𝐹(𝑥). Thus
𝑑𝑓
6
2
= 3𝑥 2 𝐹 ′ (𝑥 3 ) − 𝐹 ′ (𝑥) = 3𝑥 2 𝑒 −𝑥 − 𝑒 −𝑥
𝑑𝑥
𝐵
⃗⃗⃗⃗⃗
⃗⃗⃗⃗
2. ∫𝐴 ∇𝑓 ∙ 𝑑𝑙 = 𝑓(𝐵) − 𝑓(𝐴) = −1
𝜕𝑓
3. The problem is to calculate the function f from its gradient. Given 𝜕𝑥 = 4𝑥 3 𝑦 3 + 3 we have 𝑓 =
𝑥 4 𝑦 3 + 3𝑥 + 𝐶(𝑦), where C(y) is a function of y alone. Similarly,
𝜕𝑓
𝜕𝑦
= 3𝑥 4 𝑦 2 + 1 gives 𝑓 =
𝑥 4 𝑦 3 + 𝑦 + 𝐷(𝑥) with D(x) being a function of x alone. Comparing the two expressions for
f(x,y) we find that apart from an unimportant constant, the function f is given by 𝑓(𝑥, 𝑦) =
𝑥 4 𝑦 3 + 3𝑥 + 𝑦. Thus using the fundamental theorem of calculus, the value of the integral is -2.
4. Find the gradient of the given function and take its scalar product with the unit vector along the
given direction. Answer 6√3.
5. The normal is in the direction of the gradient at the point. Calculate the gradient and divide by
its modulus. Answer (𝑖̂ − 2 𝑗̂ + 2𝑘̂ )/3
6. In Cartesian ∇𝑓 = 2𝑖̂𝑥 − 2𝑗̂𝑦. In spherical system 𝑓(𝑟, 𝜃) = 𝑟 2 sin2 𝜃 cos 2𝜑. Use the
𝜕𝑓
1 𝜕𝑓
1
𝜕𝑓
expression for gradient ∇𝑓 = 𝑟̂ +
𝜃̂ +
𝜑̂ to calculate the gradient. This is given
𝜕𝑟
𝑟 𝜕𝜃
𝑟 sin 𝜃 𝜕𝜑
by
∇𝑓 = 2𝑟 sin2 𝜃 cos 2𝜑 𝑟̂ + 2𝑟 sin 𝜃 cos 𝜃 cos 2𝜑 𝜃̂ − 2𝑟 sin 𝜃 sin 2𝜑𝜑̂
= 2𝑟 sin 𝜃[cos 2 𝜑(sin 𝜃 𝑟̂ + cos 𝜃 𝜃̂) − sin2 𝜑(sin 𝜃 𝑟̂ + cos 𝜃 𝜃̂) − 2 sin 𝜑 cos 𝜑 𝜑̂]
= 2𝑟 sin 𝜃 cos 𝜑 (sin 𝜃 cos 𝜑 𝑟̂ + cos 𝜃 cos 𝜑 𝜃̂ − sin 𝜑 𝜑̂)
− 2𝑟 sin 𝜃 sin 𝜑 (sin 𝜃 sin 𝜑 𝑟̂ + cos 𝜃 sin 𝜑 𝜃̂ + cos 𝜑 𝜑̂)
= 2𝑥𝑖̂ − 2𝑦𝑗̂
2𝜋
7. ⃗∇⃗ ∙ 𝐹⃗ = 3. Thus ∫𝑉 ⃗∇⃗ ∙ 𝐹⃗ 𝑑𝑉 = 3 × 3 𝑅 3 = 2𝜋𝑅 3. Next step is to calculate the surface integral.
There are two surfaces, the curved surface whose outward normal is 𝑑𝑆⃗ = 𝑅 2 sin 𝜃 𝑑𝜃𝑑𝜙𝑟̂ is in
the radially outward direction and the normal to the cap which is along the negative z-direction
– 𝑘̂. For the curved surface, the surface integral is calculated easily by going over to the
spherical polar coordinates (you could also do it in Cartesian by expressing the normal in the
Cartesian using the equation to the surface). Since the normal is along the radial direction, in
computing the surface integral, we are only interested in the radial component of the force
field. It can be shown that the radial component of F is given by 𝑅 3 sin2 𝜃 + 𝑅 3 cos2 𝜃 −
𝑅 2 cos 𝜃, i.e. it is independent of 𝜑.The integral over the azimuth gives 2 𝜋 The surface integral
is
𝜋
2
2𝜋 ∫ (𝑅 3 sin3 𝜃 + 𝑅 3 cos2 𝜃 sin 𝜃 − 𝑅 2 sin 𝜃 cos 𝜃) 𝑑𝜃
0
4𝜋 3 2𝜋 3
=
𝑅 +
𝑅 − 𝜋𝑅 2 = 2𝜋𝑅 3 − 𝜋𝑅 2
3
3
That leaves us with the integral over the bottom cap where z=0. Since the normal is along the
negative z direction, the surface integral is ∫𝑆(1 − 𝑧)𝑑𝑆 = ∫ 𝑑𝑆 = 𝜋𝑅 2 . Adding the two
contributions to the surface integral, the result follows.
8. ⃗∇⃗ ∙ 𝐹⃗ = 10𝑦, ∬ 𝐹⃗ ∙ 𝑑𝑆⃗ = ∭ 10𝑦𝑑𝑉. We need to specify the limits of the volume integral. Limits
on y are from z to 4-z. Limits on z are from 0 to 2 − 𝑥 2 /2 and that of x are from 0 to 2.
2
2−𝑥 2 /2
5 ∫0 𝑑𝑥 ∫0
4−𝑧
𝑑𝑧 ∫𝑧
𝑦𝑑𝑦=128
2
9. Divergence of 𝐹⃗ is 𝑧 2 − 2𝑧. Thus the volume integral (in cylindrical coordinates) is𝜋22 ∫1 2(𝑧 −
1)𝑑𝑧 = 4𝜋.
10. We can use the use the divergence theorem to calculate the flux from the surface of the
“closed” cone first and then subtract the surface integral of the field from the to cap, at z=4. For
the closed cone, it is convenient to use spherical polar as 𝐹⃗ = 𝑟⃗, so that div F = 3. The volume
1
1
integral of div F is 3 times the volume V of the cone, i.e. 𝑉 = 3 𝜋𝑟 2 ℎ = 3 𝜋(16) ∙ 4 =
64𝜋
.
3
The
surface integral, therefore, is equal to 64𝜋. We have to now subtract the surface integral over
the top surface, whose normal is along the positive z direction. Thus 𝐹⃗ ∙ 𝑑𝑆⃗ = 𝑧𝑑𝑆. Since z=4 on
the cap and is constant the surface integral is 4 times the area of the cap which is 16π, so that
the surface integral on the cap is 64π. Thus the surface integral from the outside surface of the
open cone is zero.
11. Ans. n=1
12. (a) 0, (b) 2/r (c) 0 (d) 6
Corrections and additions to Lecture 4 :
Page 4 :
Line 7 , 10 and 11 : remove underlines from 𝑑𝑟⃗ (four places total)
Page 5 :
Line 5 :ADD after the sentence ending with “dT above”
This is known as the Fundamental Theorem of Calculus.
INSERT at the end of Page 5 :
⃗⃗𝑇 ∙ 𝑑𝑟⃗ by Δs, the length of the vector 𝑑𝑟⃗, we get, on
If we divide the relation ∆𝑇 = ∇
taking the limit,
𝑑𝑇
𝑑𝑟⃗
⃗⃗𝑇 ∙
=∇
𝑑𝑠
𝑑𝑠
𝑑𝑟⃗
However, 𝑢
⃗⃗ = 𝑑𝑠 is a unit vector along the direction in which the change of T is measured. ⃗∇⃗𝑇 ∙
𝑢
⃗⃗is known as the directional derivative of T in the direction 𝑢
⃗⃗ , and is denoted by 𝐷𝑢 𝑇.
If f(x,y,z) is a function of (x,y,z), the equation 𝑓(𝑥, 𝑦, 𝑧) = 𝑐=constant is known as the equation
to the level surface of f, parameterized by c. Suppose, 𝑟⃗(𝑡) is a curve that lies on such a level
𝑑𝑓
⃗⃗𝑓 ∙ 𝑢
surface. Then = 0. But, the left hand side is equal to ∇
⃗⃗, where 𝑢
⃗⃗ = (𝑥 ′ (𝑡), 𝑦 ′ (𝑡), 𝑧 ′ (𝑡)),
𝑑𝑡
with t being a parameter such as time. However, u is a tangent to the surface as it is a tangent to
a curve on the surface. Thus ⃗∇⃗𝑓 is along the normal to the level surface. (It makes sense as df=0
implies no change in the value of f. Thus the direction should be normal to the level surface.)
Lecture 5
Problems :
1. Calculate the curl of the following vector fields
a. 𝐹⃗ = 𝑥𝑧𝑖̂ + 𝑦𝑧𝑗̂ + 𝑥𝑦𝑘̂ (in Cartesian)
1
b. 𝐹⃗ = 𝜌 𝜌̂ (in cylindrical)
1
c. 𝐹⃗ = 𝜌 𝜃̂ (in cylindrical)
⃗⃗⃗⃗ = 𝑥𝑖̂ + 𝑥𝑗̂ + 𝑦𝑘̂, verify Stoke’s theorem over the surface of a hemisphere 𝑥 2 + 𝑦 2 +
2. For 𝐹
𝑧 2 = 𝑎2 ; 𝑧 ≥ 0 .
3. If 𝑎⃗ is a constant vector, show that ∇ × (𝑎⃗ × 𝑟⃗) = 2𝑎⃗.
4. Find the line integral of the vector field ⃗⃗⃗⃗
𝐹 = 𝑥 2 𝑖̂ + 2𝑥𝑗̂ + 𝑧 2 𝑘̂ along an ellipse in the x-y
plane 4𝑥 2 + 𝑦 2 = 4 taken in the anticlockwise direction.
5. Verify Stoke’s theorem over the boundary of a plane 𝑥 + 𝑦 + 𝑧 = 1 for the vector field ⃗⃗⃗⃗
𝐹 =
2̂
𝑦𝑖̂ + 𝑥𝑧𝑗̂ + 𝑥 𝑘.
̂ with α being a constant, to polar coordinates and calculate its
⃗⃗⃗⃗ = 𝑟 𝛼 (−𝑦𝑖̂ + 𝑥𝑗)
6. Convert 𝐹
line integral over a circle 𝑥 2 + 𝑦 2 = 𝑎2 .
⃗⃗⃗⃗ = 𝑥𝑖̂ + 𝑥 3 𝑦 2 𝑗̂ + 𝑧𝑘̂ over the boundary of
7. Using Stoke’s theorem, find the line integral of 𝐹
the semi-ellipsoid 𝑧 = √4 − 4𝑥 2 − 𝑦 2 in the plane z=0. Use both the elliptical disk and the
ellipsoidal surface to obtain the result.
8. Are the following force fields conservative?
̂
a. ⃗𝑭⃗ = 𝒙𝒛𝒊̂ + 𝒙𝒚𝒛𝒋̂ − 𝒚𝟐 𝒌
̂
⃗⃗ = 𝒛𝒊̂ + 𝒙𝒋̂ + 𝒚𝒌
b. 𝑭
𝟑
̂ − 𝟑𝒛𝟐 𝒌
̂
⃗⃗ = 𝟐𝒙𝒚 𝒊̂ + (𝟏 + 𝟐𝒙𝟐 𝒚𝟐 )𝒋
c. 𝑭
9. Calculate the Laplacian of 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑦 2 in Cartesian and in spherical coordinates and
show that the results are the same.
10. Calculate the Laplacian of the following scalar functions where 𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
a. 𝑟(cos 𝜃 + sin 𝜃 cos 𝜑)
b. 𝑟 𝑛
c.
(𝑦−𝑧)
𝑟
d. sin 𝑟/𝑟
11. A spherical shell of radius R has a total charge Q. If the charge density is given by 𝜌(𝑟⃗) =
𝐶𝛿(𝑟 − 𝑅), find the constant C.
12. Show that for 𝑎 ≠ 0, 𝛿(𝑥/𝑎) = |𝑎|𝛿(𝑥).
∞
13. Show that the derivative of the delta function satisfies ∫−∞ 𝛿 ′ (𝑥)𝑓(𝑥)𝑑𝑥 = −𝑓 ′ (0).
14. Verify the following identities
⃗⃗ × (𝑓𝐹⃗ ) = 𝑓(∇
⃗⃗ × 𝐹⃗ ) + (∇𝑓) × 𝐹⃗
a. ∇
⃗⃗ × 𝐹⃗ + ⃗∇⃗𝑓) = ⃗∇⃗ × (∇
⃗⃗ × 𝐹⃗ )
b. ⃗∇⃗ × (∇
⃗⃗ × 𝐹⃗ ) − 𝐹 ∙ (∇
⃗⃗ × ⃗G⃗)
c. ⃗∇⃗ ∙ (𝐹⃗ × 𝐺⃗ ) = ⃗G⃗ ∙ (∇
Hints for solutions to problems :
1. (a) (𝑥 − 𝑦)(𝑖̂ + 𝑗̂)
(b) 0 (only non-zero component is 𝐹𝜌 which depends on ρ (c) 0
2. First calculate the line integral ∫ 𝐹⃗ ∙ 𝑑𝑙⃗ = ∫ 𝑥𝑑𝑥 + ∫ 𝑥𝑑𝑦 + ∫ 𝑦𝑑𝑧 over a circle in z=0
plane. Parameterize the remaining integrals by 𝑥 = cos 𝑡, 𝑦 = sin 𝑡; 0 ≤ 𝑡 ≤ 2𝜋. Only
the value of the second integral is non-zero and it gives 𝜋𝑎2 .
To calculate the surface
integral, first calculate the curl
which is 𝑖̂ + 𝑘̂. If we take the circular
𝑛̂
disk as the bounding surface
of the curve, the normal l is along
the positive z-direction, so that
∫𝑐𝑢𝑟𝑙 𝐹 ∙ 𝑑𝑆 = ∫ 𝑑𝑆 = 𝜋𝑎2
𝑆
If, on the other hand, we take the surface to be the curved surface, the unit normal is
1
1
given by 𝑛̂ = (𝑖̂𝑥 + 𝑗̂𝑦 + 𝑘̂ 𝑧) . The surface integral is ∫ (𝑥 + 𝑧)𝑑𝑆. By symmetry, the
𝑎
𝑎 𝑆
term with x is zero (this can be seen in spherical coordinates, where the φ integral
vanishes). Going over to spherical coordinates, the integral is
2𝜋
𝜋/2
∫0 𝑑𝜙 ∫0
acos 𝜃𝑎2 sin 𝜃𝑑𝜃 = 𝜋𝑎2 .
3. Let the components of 𝑎⃗ be𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 . Using this write 𝑎⃗ × 𝑟⃗ explicitly and then take the
curl. Reconstruct the result.
4. Curl of the given field is 2𝑘̂. Thus the required line integral is equal to the surface
integral ∫𝑆 2𝑑𝑆 = 2∫ 𝑑𝑥 𝑑𝑦, which is twice the area of the ellipse. As the given ellipse
has a semi-major axis of a=2 and semi-minor axis of b=1, the area is 𝜋𝑎𝑏 = 2𝜋. Thus the
integral is 4𝜋.
̂
𝑖̂+𝑗̂ +𝑘
5. Curl of the vector field is – 𝑖̂𝑥 − 2𝑗̂𝑥 + 𝑘̂ (𝑧 − 1). The normal to the plane is 3 . Thus
√
the surface integral is
1
1−𝑥
1
∫(−3𝑥 + (𝑧 − 1)) 𝑑𝑆 = ∬(−3𝑥 + (𝑧 − 1))𝑑𝑥𝑑𝑦 = ∫ 𝑑𝑥 ∫ (−4𝑥 − 𝑦)𝑑𝑦
√3 𝑆
0
0
5
=−
6
2𝜋
6. In cylindrical coordinates the field is 𝐹⃗ = 𝑟 𝛼+1 𝜃̂. The line integral is ∫0 𝑎𝛼+1 𝑎𝑑𝜃 =
2𝜋𝑎2+𝛼 . Try doing it using Stoke’s theorem as well.
7. The curl of the vector is 3𝑥 2 𝑦 2 𝑘̂. The line integral of the field is equal to the surface
integral over the semi-ellipsoid , which is also equal to the surface integral of the planar
ellipse. Considering the planar ellipse, the direction of the surface is along the positive z
direction, so that,
+1
⃗⃗ × 𝐹⃗ ) ∙ 𝑑𝑆⃗ = ∫
∫(∇
𝑆
2√1−𝑥 2
3𝑥 2 𝑦 2 𝑑𝑥𝑑𝑦 = 𝜋
∫
−1
−2√1−𝑥 2
If instead, we consider the semi-ellipsoid, the normal (calculated by computing the
gradient) is 𝑛̂ =
̂
4𝑥𝑖̂+𝑦𝑗̂ +𝑧𝑘
2√16𝑥 2 +𝑦2 +𝑧 2
. Thus the integral is ∫𝑆
3𝑥 2 𝑦 2 𝑧
2√16𝑥 2 +𝑦 2 +𝑧 2
3
2
𝑑𝑆 = ∫𝑆 𝑦 2 𝑧𝑑𝑥𝑑𝑦 .
To evaluate this integral , use a coordinate transformation 𝑥 = rsin 𝜃 cos 𝜑 , 𝑦 =
2 rsin 𝜃 sin 𝜑 , 𝑧 = 2𝑟 cos 𝜃. Using this the equation to the surface becomes 4𝑟 2 =
4, 𝑖. 𝑒. 𝑟 = 1. The Jacobian of the transformation is 4𝑟 2 sin 𝜃 = 4 sin 𝜃. The surface
integral becomes
𝜋
2
2𝜋
24 ∫ sin5 𝜃 cos 𝜃 𝑑𝜃 ∫ sin2 𝜑 cos 2 𝜑𝑑𝜑 = 𝜋
0
0
8. Check if the curl is zero. Answer (a) No (b) No (c) yes
9. 𝑦 2 − 𝑥 2 = 𝑟 2 sin2 𝜃 cos 2𝜑. Laplacian is zero.
10. (a) 0 (b) 𝑛(𝑛 + 1)𝑟 𝑛−2 (c) −
2(𝑦−𝑧)
𝑟3
(d) –
sin 𝑟
𝑟
∞
11. Since the total charge is Q, ∫ 𝜌(𝑟⃗)𝑑𝑉 = 𝑄 ⇒ 4𝜋 ∫0 𝐶𝛿(𝑟 − 𝑅)𝑟 2 𝑑𝑟 = 𝑄, which gives
𝐶 = 𝑄/4𝜋𝑟 2 .
12. Take a to be positive and negative to arrive at the identity.
13. Do an integration by parts.
14. Express the vectors in terms of their components and prove each of the identities.
Corrections and additions to Lecture 5 :
1. Page 2 : Left side of the last equation should be ∮𝐶 𝐹⃗ ∙ 𝑑𝑙⃗
𝜕𝑉
𝜕2 𝑉
2. Page 10 : Last line of the equation on the right replace 𝜕𝑦2 by 𝜕𝑦2
3. Page 15 : Replace the last equation by the following equation
1
1
∫ ∇2 ( ) 𝑑 3 𝑟 = ∫ ∇ ⋅ ∇ ( ) 𝑑 3 𝑟
𝑟
𝑟
𝑉
𝑉
1
1
⃗⃗ ( ) ⋅ 𝑑𝑆⃗ = − ∫ 𝑟̂ ⋅ 𝑑𝑆⃗
= ∫∇
2
𝑟
𝑆
𝑆𝑟
MODULE :2
Lecture 6
Multiple Choice Questions :
1. Eight charges, each +q, are located symmetrically on a circle of radius R with P as its
centre. If the charge at the position X is removed and brought to the location P, the
force on this charge will be
a.
b.
c.
d.
1 𝑞2
4𝜋𝜖 𝑅 2
1 𝑞2
4𝜋𝜖 𝑅 2
7 𝑞2
4𝜋𝜖 𝑅 2
7 𝑞2
4𝜋𝜖 𝑅 2
from X to P
X
from P to X
from X to P
P
from P to X
2. Twelve charges are positioned on the dial of a wall clock such that a charge Q is at the
position 1, 2Q at position 2, 3Q at 3 and so on, ending with 12Q at the position 12. If O is
at the centre of the dial, what is the force exerted on a unit charge located at O,
assuming the radius of the dial to be of unit length?
y
a. (12 + 6√3)𝑖̂ + 6𝑗̂
b. −(12 + 6√3)𝑖̂ + 6𝑗̂
c. (12 + 6√3)𝑖̂ − 6𝑗̂
d. −(12 + 6√3)𝑖̂ − 6𝑗̂
x
3. Two small conducting spheres attract each other electrostatically. It can be concluded
that
a. At least one of the spheres is charged
b. Both the spheres are charged
c. Both the spheres are charged and their charges are of opposite sign
d. No definite conclusion on their charge state can be made from the given data.
4. Electric field lines are
5.
6.
7.
8.
a. Vectors in the direction of the electric force that acts on a test charge
b. Trajectories of a test charge in the electric field
c. Closed loops
d. Pictorial representation of electric field around a charged object.
The total amount of negative charge of all the electrons contained in one mole of water
is approximately
a. 1C
b. 105 C
c. 106C
d. 1023C
A negative charge of 9C and mass 2kg orbits around a heavy positive charge of 16C
in a circular orbit of radius 5m. What is the speed of the negative charge?
a. 180 m/s
b. 324 m/s
c. 360 m/s
d. 1984 m/s
A 2.5 C test charge is placed to the right of another charge Q. If there is an attractive
force of 22.5 N between the two, what would be the force exerted if the magnitude of
the test charge were to be doubled but it stayed at the same location as before?
a. 90 N
b. 45 N
c. 11.25 N
d. Depends on the distance between the test charge and the charge Q.
The charges 𝑄, 𝑄, −4𝑄and − 𝑄 are kept at the corners A, B, C and D , respectively, of a
square. If the force between the charges at A and B is F, the net force exerted on the
charge at A due to the other three charges is
A
B
e.
a. 6F
b. 3𝐹
c. √6𝐹
d. √3𝐹
9. Two small conducting spheres of mass m and charge q each are suspended from a
common point by means of threads. The spheres settle down to an equilibrium position,
C
D
each making an angle  with the vertical. The tension in either of the threads is
a. Zero
b. Greater than mg
c. Less than mg
d. Equal to mg
10. Two opposite charges are placed on the paper, as shown in the figure.
The charge on the left is three times as big as the charge on the right. Other than at infinity,
where else can the force on a unit test charge due to these two charges is zero?
a. To the right of the smaller
charge
b. Between the two charges
c. To the left of the bigger
charge
d. Depends on the sign of the
test charge.
d
Problems :
1. ABC is an equilateral triangle of side 40 cm. At the vertices A and B +4C of charge is
kept fixed while at the vertex C a −4𝜇C charge is held. What is the force on the charge
at C? If now, the charge at C is released while the charges at A and B still remain fixed,
describe its subsequent motion.
A
C
B
2. Two equal and positive charges, q each, are at a finite distance 2d from each other. A
third charge Q is located at the midpoint of the line joining the two. Where should a unit
positive charge be placed so that the net force on it zero? What, if any, is the
requirement of the magnitude and sign of the charge Q?
3. Twelve equal charges +q are situated on a circle of radius R and they are equally spaced
like the position of digits on the dial of a clock. What is the net force on a charge Q kept
at the centre? What would be the force on Q if the charge at 3’o clock position is
removed?
4. Two particles, each of mass m and having charges q and 2q are suspended by strings of
length l from a common point. Find the angle  that each string makes with the vertical.
5. A particle of mass m and with charge q is suspended from a peg on a wall by means of
a string of length 0.5m. The string makes an angle 600 with the vertical. Another charge
q is held at the same horizontal level as the first charge so that the distance between
the charges is R. Calculate the tension in the string and the distance R.
6. A wire is bent in the form of a semicircle and carries a linear charge density . Find the
electric field at the centre of the circle.
Answers to Multiple Choice Questions :
1. (b) 2. (c) 3. (a) 4. (d) 5. (c) 6. (c) 7. (b) 8. (c) 9. (b) 10. (a)
Hints for solutions to problems :
1. Resolve the forces along the perpendicular bisector from C on to AB and a direction
perpendicular to it. The net force is along the perpendicular bisector towards AB and
has a magnitude 1.56 N.
2. Charge Q has to have opposite sign of the charge q; else at no finite distance field can be
zero. In order that forces cancel out, the test charge must be located along the
perpendicular at Q of the line joining the two charges q.
q

Q
P
q
Resolve the forces due to the pair q, the components perpendicular to QP cancel. Along
QP, it has to be canceled by the repulsive force due to Q. The condition gives cos 3 𝜃 =
𝑄/2𝑞.
3. When equal charges are kept on the dial, the net force on a charge at the centre is zero
as the forces due to a pair kept in diametrically opposite positions cancel. If a single
charge is removed, the effect is due to the diametrically opposite member.
4. Draw the freebody diagram of each particle. Resole forces along vertical and horizontal.

T
Fe
2q
mg
q
𝑻 𝐜𝐨𝐬 𝜽 = 𝒎𝒈

𝑻 𝐬𝐢𝐧 𝜽 =
𝟏
𝟐𝒒𝟐
𝟒𝝅𝝐𝟎 𝟒𝒍𝟐 𝐭𝐚𝐧𝟐 𝜽
Eliminate T to get the angle .
5. Very similar to Problem 4. The right hand side of the second equation is to be changed
to
𝟏
𝒒𝟐
𝟒𝝅𝝐𝟎 𝑹𝟐
. Since the angle is given and T can be eliminated, one can find R.
6. Consider the field at P due to an element Rd located at an angle , as shown. The
1
magnitude of the field due to this element is given by 𝑑𝐸 = 4𝜋𝜖
𝜆𝑅 𝑑𝜃
0
𝑅2
. The component
of the field parallel to the diameter cancels due to a symmetrically placed element,
leaving the net field to be perpendicular to the diameter and pointing away from the
semicircle.

P
𝜋
1
The net field has a magnitude |𝐸| = 4𝜋𝜖 2 ∫02
0
𝜆𝑅 𝑑𝜃
𝑅2
𝜆
sin 𝜃 = 2𝜋𝜖
0𝑅
Corrections to Lecture 6:
Page 1 : line 8 : funadental fundamental
Page 6 : Align mathematical expressions on lines 2, 6, 8 with text
Page 7 : Align mathematical expression on lines .
In the last but one line of the last equation in the second term on the right insert 𝑟̂2
Page 8 : align mathematical expressions with text.
Line 1 : Field  field
Line 4 : Electric Field Electric Field
Line 3 from bottom :colobm coulomb
Page 9 : align math with text
last line : theirdensity their density
Line 3 from bottom :auniformly a uniformly
Lecture 7
Multiple Choice Questions :
1. The figure shows field lines due to three electric charges each of magnitude Q. The red circle
indicates the intersection of a spherical Gaussian surface with the plane of the paper. The net
electric flux out of the sphere is
a. 2𝑄/𝜖0
b. −2𝑄/𝜖0
c. zero
d. cannot be determined without knowing which charge is negative
2. The figure shows a pyramid with a square base. Each of the triangular side is an equilateral
triangle. A charge Q is fixed at the centre of the base of the pyramid. The flux through any of
the triangular sides is
a. 𝑄/𝜖0
b. 𝑄/2𝜖0
c. 𝑄/4𝜖0
d. 𝑄/8𝜖0
3. The field in the cube of side L is given by 𝐸⃗⃗ = (𝑎 + 𝑏𝑥)𝑖̂, where a and b are constants.
The total charge enclosed within the cube is
z
a.
b.
c.
d.
Zero
aL3
bL3
(a+b/2)L3
O
x
y
4. A point charge Q is situated at a distance d/2 directly above a square plate of side d. The
electric field through the square is
a.
b.
c.
d.
𝑄/6𝜖0
𝑄/4𝜖0
𝑄/3𝜖0
𝑄/𝜖0
d/2
d
d
5. A constant electric field 𝐸⃗⃗ passes through the hemispherical shell of radius R. The flux
through the curved surface is
a.
b.
c.
d.
Zero
𝐸𝜋𝑅 2
2𝐸𝜋𝑅 2
4𝐸𝜋𝑅 2
𝐸⃗⃗
Problems :
1. The electric field corresponding to a charge distribution is given by
𝑎
𝑎2
𝑟
⃗
−
𝑟⃗ for 𝑟 < 𝑅
⃗⃗
𝐸 = {𝑟 3
𝑅3
0
𝑟≥𝑅
Where a is a real constant. Find the charge distribution (charge density and
the total charge) which gives rise to this field. Does your answer depend on
whether a is positive or negative?
2. The field in the hemisphere of radius R is given by 𝐸⃗⃗ = (𝑎 + 𝑏𝑥)𝑖̂, where a
and b are constants. What is the total charge within the hemisphere?
3. Four spheres of equal radii R=2a are placed such that their centres are at the
corners of a side of a square ABCD of side a. The spheres, with their centres
at A,B,C and D have charge densities +𝜌, −𝜌, +𝜌 and – 𝜌 respectively. Ahow
that in the region of space where all the spheres overlap, the electric field
vanishes.
A
B
D
C
4. A 5.31mC charge is located at one corner of a cube of side 8cm. Calculate the
flux of electric field through one face of the cube.
5. A sphere of radius R has an off-centre spherical hole of radius R/4, the centre C of
the hole being at a distance R/2 from the centre O of the sphere. The sphere has a
charge density . Obtain an expression for the electric field at an arbitrary point of
the hole.
O
6. A plate has
C
a thickness d along the z direction and is of infinite extent in
R
𝑑
2
𝑑
2
the x-y plane. The volume charge density is for – ≤ 𝑧 ≤ + . Calculate the
electric field for all values of z.
7. A cylinder has a circular cross-section radius R and is of infinite length. It has
a volume charge density . Calculate the electric field at all distances r from
the axis of the cylinder.
Answers to Multiple choice questions:
1. (c) 2. (d) 3. (c) 4. (a) 5. (b)
Hints for solutions to Problems
1. The first term represents the electric field due to a point charge 𝑄 = 4𝜋𝜖0 𝑎 at the origin
while the second term represents the field due to a uniformly charged sphere with a
total charge −4𝜋𝜖0 𝑎2, i.e. having a negative charge density which is 𝜌 = −3𝜖0 𝑎2 /𝑅 3 .
This can be checked by taking the divergence of the given electric field. For 𝑎 > 0 , the
point charge is positive while for 𝑎 < 0 it is negative. Note however, that the field
vanishes at distances greater than R. Thus if we take a Gaussian sphere (the field is
spherically symmetric) for r>R, the surface integral being zero, the net charge enclosed
should be zero. This is only possible if a>0.
3
⃗⃗ = 𝒃= constant. Thus the total charge is 2𝜋𝑅 𝑏. Try doing this problem by
2. 𝝆 = ⃗𝛁⃗ ⋅ ⃗𝑬
3
calculating the surface integral over the curved surface (the surface integral over the
circular cap is zero).
3. First show that the field in the region of intersection of two identical spheres which
carry opposite charge is constant and is directed along the line joining the two centres.
P
O1
O2
𝜌
⃗⃗⃗⃗⃗⃗⃗⃗
The field at P due to the sphere with centre at O1 is (3𝜖 )𝑂
1 𝑃 while that due to the sphere
0
𝜌
𝜌
⃗⃗⃗⃗⃗⃗⃗⃗2 , so that the net field is ( ) 𝑂
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
with centre at O2 is (3𝜖 )𝑃𝑂
1 𝑂2 . Using this it is easy to show
3𝜖
0
the given result.
0
4. Imagine stacking eight identical cubes of side a=8cm so that we get one single cube of
side 2a.
The charge can be made to reside at the centre of this big cube. Because of the symmetry,
the flux through each face of this big cube is 𝑄/6𝜖0 . Since each face of the big cube is one
fourth the size of a face of the original cube, the flux through each face of the original is
𝑄
24𝜖0
.
5. The hole can be considered as a superposition of a sphere with charge density with a
sphere of identical size but with charge density –. Consider a point P in the hole. The
field at P is a superposition of field due to the big sphere with a charge density 
everywhere and the field due to a negatively charged sphere of density  of the size of
𝜌
𝜌
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
the hole. The field due to former is
𝑂𝑃 while that due to the latter is
𝑃𝐶 , which
3𝜖0
3𝜖0
𝜌
⃗⃗⃗⃗⃗⃗ , an expression independent of the position of P inside the hole.
gives 3𝜖 𝑂𝐶
0
6. The field can only depend on the z-distance from the centre plane of the slab. The
Gaussian pillbox is in the shape of a rectangular parallelepiped.
𝜌
𝑑
2𝜖0
E
-d/2
d/2
-d/2
−𝜌
𝑑
2𝜖0
x
x
O
x
d/2
Since the field directions are perpendicular to the slab, the flux contribution is from the
two faces perpendicular to z direction. If the area is taken to be A, the magnitude of the
𝑄
1
0
0
𝜌
field is given by 2|𝐸|𝐴 = 𝜖 = 𝜖 𝐴. 𝑑. 𝜌, which gives |𝐸| = 2𝜖 𝑑 for x>d/2.
0
𝑄
If x<d/2, only a part of the charge is included. In this case we have, 2|𝐸|𝐴 = 𝜖 =
0
1
𝜖0
𝜌
𝐴. 2𝑥. 𝜌, which gives |𝐸| = 𝜖 𝑥. The field distribution is shown above.
0
7. Choose the Gaussian surface to be a coaxial cylinder of length L and radius r. Field lines
being perpendicular to the cylinder, contribution to flux is from the curved surface
alone.
Flux = |𝐸|2𝜋𝑟 𝐿 . The amount of charge enclosed is=
𝜌𝜋𝑅 2 𝐿, for 𝑟 > 𝑅 an𝑑 = 𝜌𝜋𝑟 2 𝐿 for 𝑟 < 𝑅
𝜌𝑅 2
Thus the magnitude of the field is 2𝜖 𝑟 for 𝑟 > 𝑅 and
0
𝜌𝑟
2𝜖0
for 𝑟 < 𝑅 .
CORRECTIONS
There are major corrections in this lecture some pages are mixed up.
Page 3 : Align line 2
Page 4 : Example  Example 1 (This is to be corrected inside the link as well)
Exercise  Exercise 1
Insert at the bottom : A linked item : Example 2
The linked page should be as follows ;
Example 2 :
A flat surface with an area of 0.2 m2 lies in the xy plane in a uniform electric field
given by 𝐸⃗⃗ = 5𝑖̂ + 3𝑗̂ + 3𝑘̂ (N/C). Find the flux through the surface.
Solution : Since the surface is in the xy plane, the surface vector is 0.2 𝑘̂ (m2).
Thus the flux = 𝐸⃗⃗ ∙ 𝑆⃗ = 0.6 Nm2/C.
Page 6 : Line 4 from bottom : align 
Page 7 : Replace the page as follows :
Field due to an infinite line charge with charge density 
Gaussian surface is a cylinder of radius rand length L whose axis is along the line
charge.
Normal to the Gaussian surface is everywhere perpendicular to the line charge.
By symmetry, the magnitude of the electric field is the same everywhere on the
Gaussian surface and its direction is along the outward normal. Hence
∫ 𝐸⃗⃗ ∙ 𝑑𝑆⃗ = |𝐸|2𝜋𝑟𝐿
The amount of charge contained inside the cylinder is 𝜆𝐿. Thus,
Exercise  Exercise 2
Delete : Example (also delete the linked page : This is repeated later).
Page 8 : Last but one line align 𝜌̂
Page 10 : Line 3 from bottom and Page 11 and 12 should come immediately
after page 8 (i.e. become pages 9 and 10)
The current page 9 and 10 (except last three lines of page 10 should become
new pages 11 and 12.
Page 13 : Recap : 1st item : normalcomponent normal component
Lecture 8
Corrections :
1. Example 11 should end at the last line of the page of the example. The
statement in bold face “Line integral …. Zero” is a part of the main text. The
rest of the pages of this example should be a part of the main text.
2. Page 6 : 0 is a subscript of P (P0)
3. Exercises should all be properly numbered. Inside the exercise numbers
should be corrected.
Multiple Choice Questions :
1. A point charge -3Q lies at the centre of a conducting shell of radius 2R.
The net charge on the outer surface of the shell is
a. -3Q
b. Zero
c. +1.5 Q
d. +3Q
2. Two identical spherical conductors A and B of radius R , each carrying a
charge Q are kept at some distance from each other. A third spherical
conductor C , initially uncharged, is first brought into contact with A and
then with B before finally being removed to a far away distance. If the
charge on C is (10/9)Q, the radius of C is
a. R/5
b. R/3
c. R/2
d. 2R
3. 64 identical spherical drops of mercury are combined to form a large
drop. If the potential of each smaller drop is 1 V, the potential of the
final drop is
a. 1 V
b. 4 V
c. 16 V
d. 64 V
4. Infinite number of charges of equal magnitude Q are placed along the
axis at distances a, 2a, 3a, ….from the origin. If the charges alternate in
sign and the charge closest to the origin is positive, the potential at the
origin due to the charge distribution is
a.
b.
c.
−𝑄
4𝜋𝜖0 𝑎
𝑄
4𝜋𝜖0 𝑎
𝑄𝑙𝑛 2
4𝜋𝜖0 𝑎
d. Zero
5. Three charges, Q, q and q are arranged at the vertices of a right angled
isosceles triangle of base a. If the charge Q are fixed, the configuration
has minimum electrostatic energy when q is equal to
a. −𝑄
b. +𝑄
√2+1
√2
√2+1
Q
Q
√2
q
q
Q
Q
c. −𝑄
√2+1
2√2
d. +𝑄
√2+1
2√2
6. The electric potential in a region along the x axis varies with distance x
(in meters) as 𝑉(𝑥) = 4 + 4𝑥 2 (Volts). The force acting on a 1C
charge located at 𝑥 = −3 is
a. 2.4 × 10−5 N along the positive x axis
b. 2.4 × 10−5 N along the negative x axis
c. 4 × 10−5 N along the positive x axis
d. 4 × 10−5 N along the negative x axis
7. Four electric charges +q, +q, –q and –q are placed at the corners of a
square of side 2L (see figure). The electric potential at point A, midway
between the two charges +q and +q, is
a.
b.
c.
𝑞
2𝜋𝜖0
𝑞
(1 +
𝐿
2𝜋𝜖0 𝐿
𝑞
2𝜋𝜖0 𝐿
(1 −
(1 +
1
)
+𝑞
−𝑞
√5
1
)
√5
1
)
√3
d. Zero
−𝑞
+𝑞
8. Two uniformly charged concentric rings of radii R and 2R are placed on
a plane. Each ring has a charge density The electric potential at the
centre of the rings is
a.
b.
c.
d.
2𝜆
3𝑅𝜖0
𝜆
𝑅𝜖0
𝜆
𝜖0
2𝜆
3𝜖0
9. A hemisphere of radius R is charged with a uniform surface charged
density s on its curved surface. The potential the centre is
a.
b.
c.
d.
𝜎
4𝜖0
𝜎
2𝜖0
𝜎
𝜖0
4𝜎
3𝜖0
10. A hollow metal sphere of radius R is charged to a potential of 10 V on
its surface. What is the potential at the centre of the sphere?
a.
b.
c.
d.
Zero
+10 V
-10 V
Same as its value outside at a distance of R from the surface of
the sphere.
Problems
I
II
III
1. A uniformly charged sphere of radius R contains a charge Q.
Choose the origin of coordinate system at the centre of the
sphere and let V(0)=0. Find an expression for the potential both
inside and outside the sphere.
2. Two identical metal plates have area 1 m2 each and are separated
by 3 cm. Initially both are uncharged. A charge of 2 nC is
transferred from the plate on the left to the plate on the right and
equilibrium is allowed to be established. Neglecting edge effects,
calculate the electric field (a) in the region II between the plates
at a distance of 0.5 cm from the plate to the right and (b) at
points to the immediate left of the plate on the left (region I) and
at points to the immediate right of the plate to the right in region
III.
3. Two identical thin rings, each of radius R, are placed coaxially at
a distance R from each other. The rings carry charges Q1 and Q2
uniformly spread over the rings. Find the work done in removing a
charge q from the centre of the first ring to that of the second
ring.
4. A metal sphere of radius R carrying a charge q is surrounded by a
thick concentric conducting shell of radii a and b. The shell has no
charge. Calculate the charge densities on the metal sphere as well
as on the inside and outside surface of the shell. With the
reference of the potential at infinite distances, determine the
potential at thecentre of the sphere. If now, the outer surface of
the shell is grounded, what will be the potential at the centre of
the sphere?
5. A metal sphere of radius R has two spherical cavities of radii a
andb. The former has a charge q at the centre of the cavity while
the latter has a charge Q at its centre. Find the charge densities in
the interior of the two cavities and on the outside surface of the
6.
7.
8.
9.
sphere. Also determine the field outside the sphere at a distance r
from the centre of the sphere.
Potential on the surface of a sphere is 400 V and its value drops to
100 V at a distance of 60 cm from the surface. Calculate the
radius of the sphere and the amount of charge contained in the
sphere.
Two equal charges q are located at a distance 2d apart. Find the
expression for the electric potential at a distance of z along the
perpendicular bisector of the line joining the two charges. Using
this determine the electric field at that point.
A metal sphere of radius R carries a charge Q. Surrounding this is a
concentric metallic shell of inner radius 2R and outer radius 3R. If
the shell carries a total charge 3Q, find the potential for r<R.
Can an electric field be given by the expression 𝐸⃗⃗ = 3𝑥𝑦𝑖̂ +
2𝑦𝑧𝑗̂ + 𝑥𝑧𝑘̂ ?
10. How much work is done in moving a 0.5nC charge on the surface
of a sphere of radius 2m from an initial position 𝜃 = 0, 𝜑 = 0 to a
final position 𝜃 = 0, 𝜑 =
𝜋
2
? The electrostatic field in the region
is given by 𝐸⃗⃗ = 10𝑦𝑖̂ + 10𝑥𝑗̂ + 2𝑧𝑘̂ (in V/m).
Answers to Multiple choice questions:
1. (a) 2. (d) 3. (c) 4. (c) 5. (c) 6. (d) 7. (b) 8. (c) (9) (b) 10. (b)
Hints for solutions to Problems
𝑄
1. Electric field is known from Gauss’s law to be 4𝜋𝜖
𝑟 𝑄
,𝑉(𝑟) − 𝑉(0) = − ∫ 𝐸⃗⃗ ∙ ⃗⃗⃗⃗
𝑑𝑙 = − ∫0 4𝜋𝜖
𝑟
0
𝑅3
𝑟𝑟̂
3
0𝑅
𝑟2
𝑄
𝑑𝑟 = − 8𝜋𝜖
0
𝑅3
𝑄
for r <R and 4𝜋𝜖
𝑟̂
2
0𝑟
for r>R. For r<R
. For r>R, take the corresponding
expression for the field for outside the sphere and integrate from the surface of the sphere to
𝑟
𝑄
the point where the potential is to be found. We get 𝑉(𝑟) = 𝑉(𝑅) − ∫𝑅 4𝜋𝜖
𝑄
4𝜋𝜖0 𝑟
1
2
0𝑟
3𝑄 1
𝑑𝑟 = − 8𝜋𝜖
0
𝑅
+
.
2. The charge on the left plate is -2nC and on the right plate is +2nC. Without assuming that the
final charges are on the faces of plate facing each other, we can assume that the charge density
on the left plate (plate 1) is 1L on its left face and 1R on its right face. Likewise, the charge
density on the right plate (plate 2) are 2Land 2R respectively. We also have 𝜎2 = −𝜎1 = 𝜎 =
2×10−9
1
= 2 × 10−9 C/m2. Further, 𝜎1 = 𝜎1𝐿 + 𝜎1𝑅 and𝜎2 = 𝜎2𝐿 + 𝜎2𝑅 . For a point in region
𝜎
II,𝐸𝐼𝐼 = 2𝜖1𝐿 +
𝜎1𝑅
0
2𝜖0
𝜎
𝜎
− 2𝜖2𝐿 − 2𝜖2𝑅 =
0
𝜎1 −𝜎2
0
2𝜖0
= −2𝜎/𝜖0 .
1L 1R
2L
2R
Substituting values, the field strength is approximately 226 N/C. By similar arguments, the field
to the left of the left plate and that to the right of the right plate can be shown to be zero.
(Solution becomes simpler if one assumes a- priori that the charges appear only on the faces of
the two plates that face each other.)
𝑄
3. Calculate the potential at P due to the first ring which is 𝑉1 (𝑃) = 4𝜋𝜖1
1
0
𝑅
. Similarly, since P is
at a distance √2 𝑅 from all points on the second ring, the potential at P due to the second ring
𝑄
is 𝑉2 (𝑃) = 4𝜋𝜖2
1
0
√2𝑅
1
𝑄
. Thus the net potential at P is 𝑉(𝑃) = 4𝜋𝜖 ( 𝑅1 +
0
1
𝑄
potential at P’ is 𝑉(𝑃′) = 4𝜋𝜖 ( 𝑅2 +
0
between P and P’, which is
𝑄1
√2𝑅
𝑄2
√2𝑅
). Similarly, the
). The work done is q times the potential difference
𝑞(𝑄1 −𝑄2 ) √2−1
4𝜋𝜖0 𝑅
√2
.
P
P’
4. The charge densities are 𝜎𝑅 =
𝑄
; 𝜎𝑎
4𝜋𝑅2
=−
𝑄
; 𝜎𝑏
4𝜋𝑎 2
=
𝑄
.
4𝜋𝑏2
Q1
Q2
Inside the metal sphere (for r<R) the
1
field is zero. Field for R<r<a as well as for r>b, the field is inverse square and is given by 4𝜋𝜖
potential can be calculated as a line integral of the electric field
𝑏
𝑅
1 𝑞
1 𝑞
𝑞 1 1 1
𝑉(0) = − ∫
𝑑𝑟
−
∫
𝑑𝑟 =
[ + − ]
2
2
4𝜋𝜖0 𝑏 𝑅 𝑎
∞ 4𝜋𝜖0 𝑟
𝑎 4𝜋𝜖0 𝑟
𝑞
2 𝑟̂ .
𝑟
0
The
If the outside of the shell is grounded, the charge density on that surface becomes zero while
the charge densities on all other surfaces remain unchanged. The only non-zero field is in the
𝑅
region R<r<a. Further V(b)=0. Thus 𝑉(0) = − ∫𝑎
1 𝑞
𝑑𝑟
4𝜋𝜖0 𝑟 2
=
𝑞
1
[
4𝜋𝜖0 𝑅
1
− ].
𝑎
𝑞
𝑄
5. The charge densities on the inside surface of the cavities are 𝜎𝑎 = − 4𝜋𝑎2 ; 𝜎𝑏 = − 4𝜋𝑏2 , while
𝑞+𝑄
the charge density on the outside surface is 𝜎𝑅 = 4𝜋𝑅2 . The field outside the sphere is same as due to a
charge q+Q located at the centre of the sphere.
6. R=0.2 m and 𝑄 = 8.88 × 10−9 C.
7. Take the line joining the charges along the x-axis and the perpendicular bisector along the z axis. Since
the point at which the potential is to calculated is at a distance √(𝑑2 + 𝑧 2 ) from each of the charges,
𝑞
2
. The electric field
4𝜋𝜖0 √(𝑑 2 +𝑧 2 )
𝑞 𝑑
1
𝑞
𝑧
− 2𝜋𝜖 𝑑𝑧
𝑘̂ = 2𝜋𝜖 (𝑑2 +𝑧2 )3/2 𝑘̂ .
2
2
√(𝑑 +𝑧 )
0
0
the potential is given by 𝑉(𝑧) =
given by𝐸⃗⃗ = −∇𝑉 =
is the gradient of the potential and is
8. The potential in the region r<R is the same as that on the surface of the metal sphere. The charge
distribution on the shell will be –Q on the inner surface and 4Q on the outer surface. We can use
superposition principle to find the potential at r=R. This gives 𝑉 =
1
𝑄
(
4𝜋𝜖0 𝑅
−
𝑄
2𝑅
+
4𝑄
)
3𝑅
=
11𝑄
.
24𝜋𝜖0
You can
also find it by determining the field distribution using Gauss’s law and determining potential
9. No, the curl the given field is not zero (calculate the curl explicitly).
10. Show that the field is conservative and obtain the corresponding potential. The potential is given by
𝑉 = −10𝑥𝑦 − 𝑧 2 (in V). Then convert the given positions to Cartesian. The work done is equal to q
times the change in potential. Ans. 6 × 10−9 J.
Lecture 9
Corrections :
1. Align and remove unnecessary underline on all pages.
2. Page 7 : Beginning of section B : I in In is displaced.
3. Page 12 : 1, 2, 23 are subscripts to P on this page.
Add Examples :
1. Obtain an expression for the energy of a uniformly charged sphere containing a total
charge Q.
Solution :(Method 1)
1
We will first use the expression for the energy 𝑊 = 2 ∫𝑉 𝜌𝑉𝑑 3 𝑥 where the integral is
over the volume of the sphere since the charge density outside is zero. The potential
within the sphere can be calculated from the electric field, which in turn, can be
𝑄 𝑟
𝑄 1
calculated using Gauss’s law as 𝐸⃗⃗ =
𝑟̂ for r<R and =
𝑟̂ for r>R. Taking the
3
2
4𝜋𝜖0 𝑅
4𝜋𝜖0 𝑟
zero of the potential to be at infinite distance,
𝑅
𝑟 𝑅
𝑄 1
𝑄 𝑟
𝑉(𝑟) = − ∫
𝑑𝑟
−
∫
∫
𝑑𝑟
2
3
∞ 4𝜋𝜖0 𝑟
𝑅 ∞ 4𝜋𝜖0 𝑅
𝑄
3
𝑟2
=
[ − 3]
4𝜋𝜖0 2𝑅 2𝑅
𝑄
The charge density is constant and is given by 𝜌 = 3 4𝜋𝑅3 . Thus the energy is given by
𝑄
1
[∫ 3𝜌𝑑𝜏 − 2 ∫ 𝑟 2 𝜌𝑑𝜏 ]
16𝜋𝜖0 𝑅 𝑉
𝑅 𝑉
𝑅
𝑅
1𝑄
𝑄
𝑄
2
=
[∫ 9
4𝜋𝑟
𝑑𝑟
−
∫
3
𝑟 2 4𝜋𝑟 2 𝑑𝑟]
3
3
16𝜋𝜖0 𝑅 0 4𝜋𝑅
4𝜋𝑅
0
2
2
3𝑄
1
3𝑄
=
[1 − ] =
16𝜋𝜖0 𝑅
5
20𝜋𝜖0 𝑅
𝑊=
Method 2
Here, we will take the expression for the energy to be given by the volume integral over the
square of the electric field, remembering that the expression for the field are different inside
the sphere and outside.
𝑊=
𝜖0
∫ 𝐸 2 𝑑𝜏
2 𝑉
Thus the energy is given by
𝜖0 𝑅 𝑄 2 𝑟 2
𝜖0 ∞ 𝑄 2 1
2
∫
4𝜋𝑟
𝑑𝑟
+
∫
4𝜋𝑟 2 𝑑𝑟
2 0 16𝜋 2 𝜖02 𝑅 6
2 𝑅 16𝜋 2 𝜖02 𝑟 4
∞
𝑄2 1 𝑅 4
1
=
[ 6 ∫ 𝑟 𝑑𝑟 + ∫ 2 𝑑𝑟]
8𝜋𝜖0 𝑅 0
𝑅 𝑟
𝑊=
𝑄2 1
1
3𝑄 2
=
[ + ]=
8𝜋𝜖0 5𝑅 𝑅
20𝜋𝜖0 𝑅
Method 3
Here we will take the expression for the energy to be given by the sum of a volume integral
over the sphere and a surface integral over the same,
𝜖0
𝜖0
∫ 𝐸 2 𝑑𝜏 + ∫𝑉𝐸⃗⃗ ⋅ ⃗⃗⃗⃗⃗
𝑑𝑆
2 𝑉
2 𝑆
The integrals are calculated using the expressions for the field and potential given under
Method 1. The volume contribution is the same as calculated in the first term of method 2, i.e.
𝑊=
𝑄2
40𝜋𝜖0 𝑅
. The surface terms is easily calculated since we only need the field and potential on the
surface, i.e. at r=R. The surface contribution is given by
𝜖0
𝜖0 𝑄
3
𝑅2
𝑄𝑅
∫𝑉𝐸⃗⃗ ⋅ ⃗⃗⃗⃗⃗
𝑑𝑆 = [
[ − 3] ×
] ∫ 𝑑𝑆
2 𝑆
2 4𝜋𝜖0 2𝑅 2𝑅
4𝜋𝜖0 𝑅 3
𝑄2
𝑄2
2
=
4𝜋𝑅 =
32𝜋 2 𝜖0 𝑅 3
8𝜋𝜖0 𝑅
Adding the two contribution result follows.
Problems
1. Four identical charges are at the vertices of a tetrahedron of side a. What is the amount of work
required to assemble this charge configuration?
2. Calculate the electrostatic energy of a metallic sphere of radius R containing a charge Q.
3. An isolated conducting sphere of radius 10 cm is charged to a potential of 900 V. What is the
electrostatic energy of the sphere?
4. Three point charges – 𝑞, −𝑞 and + 3𝑞 are at the vertices of an equilateral triangle of side a. A
fourth particle of charge q and mass m is initially at rest at the centre of the triangle. When the
particle is released what would be its kinetic energy when it is far away from the triangle?
5. A conducting sphere of radius R is surrounded by a concentric spherical conducting shell of
inner radiusa and outer radius b. The sphere and the shell are initially charge neutral. How
much work has to be done to transfer charge from the sphere to the shell till the inner sphere
has a charge +Q?
Hints for solutions to Problems
6𝑞2
1.
There are 6 pairs of interaction terms. The work is 4𝜋𝜖 𝑎.
2.
The electrostatic energy is the amount of work required to assemble the charge on the surface
of the sphere from an infinite distance. Since the potential on the surface of the sphere is
constant and is given by 𝑉 = 𝑞/4𝜋𝜖0 𝑅, the amount of work required to bring a charge dq from
0
𝑞𝑑𝑞
infinity to the surface when the surface has a charge q is 𝑑𝑊 = 4𝜋𝜖 𝑅. Integrating from 0 to Q,
0
the electrostatic energy is 𝑊 =
3. The potential is given by 𝑉 =
𝑄2
.
8𝜋𝜖0 𝑅
𝑄
4𝜋𝜖0 𝑅
which gives 𝑄 = 10−8 C. The energy is ½ QV which is 4.5 J
4. The initial kinetic energy is zero. The final potent1ial energy is zero. Thus the final kinetic energy
1
is equal to the initial potential energy, which is 4𝜋𝜖
𝑞2
,
0 𝑏
where 𝑏 = 𝑎/√3, is the distance of the
1
2
centre from each vertex. Equating this to 𝑚𝑣 2 , the result follows.
5. When there is a charge q on the inner sphere, the potential difference between the sphere and
𝑞
1
1
the shell is 𝑉(𝑞) = 4𝜋𝜖 (𝑅 − 𝑎). In this situation to transfer an additional charge dq from the
0
shell to the inner sphere, an amount 𝑉𝑑𝑞 has to be done. Thus the total work done is
𝑄
𝑞
1 1
𝑄2 1 1
𝑊=∫
( − ) 𝑑𝑞 =
( − )
𝑎
8𝜋𝜖0 𝑅 𝑎
0 4𝜋𝜖0 𝑅
Lecture 10
Problems
1. A 5 nC charge is located at the origin. Find the radius of the equipotential
sphere for which the potential is 100 V.
2. The figure below shows equipotential contours corresponding to a charge
distribution. Near which of the points A, B, C, D, is the electric field strength
strongest and points in the positive x direction?
5V
5V
0V
B
C
A
y
B
B
D
x
3. Using the definition of work, show that the local electric field is perpendicular to
the equipotential surface.
4. The charge density on a unit disk in the x-y plane is constant and is given by 0.
The charge density vanishes on the boundary of the disk. Obtain an expression
for the potential on the disk.
y
1
1
-1
x
-1
5. For the problem 4 above, what is the equipotential contour?
6. Determine the charge distribution that would give rise to the potential 𝜑(𝑟) =
𝜑0 𝑒 −𝜆𝑟
.
𝑟
7.
8.
9.
10.
Note that there is a singularity at r=0 and hence treat the origin
separately. What is the total charge corresponding to this distribution?
A charge +2q is located a t the point (4,0,0) and another charge –3q is located
at (9,0,0). If the charges are in Coulomb and the distances in meters, obtain the
equation of the equipotential surface on which the potential is zero.
A parallel plate capacitor of capacitance C is charged fully by a battery of emf E.
The battery is now disconnected. The plates of the capacitor are pulled apart so
that the new separation between the plates is double the original separation. By
what factor does the initial energy change by the process of pulling the plates
apart?
A cylindrical capacitor of capacitance 20 pF has an outer radius twice the size of
its inner radius. The length of the cylinder is ten times the outer radius. A charge
+Q= 10 nC is evenly distributed on the outer surface of the inner cylinder and
a charge –Q is similarly distributed on the inner surface of the outer cylinder.
Find the energy density at any point of the capacitor. Integrate this to find the
total energy stored in the capacitor.
A metallic sphere with a capacitance of 20 pF is charged to a potential of 100 V.
It is then connected to an uncharged sphere of capacitance 80 pF by means of a
long thin conducting wire. How much of initial energy is lost in the process?
Hints for solutions to Problem
𝑄
1. The potential at a distance R is given by 4𝜋𝜖
0𝑅
. Substituting the values of charge and the
potential, the radius R = 0.45 m.
2. The electric field is given by negative gradient of the potential. The field is strongest where the
equipotentials are closer. Using these two facts it follows that the point B is the required point.
3. Work is given by 𝑊 = 𝐹⃗ ∙ 𝑠⃗ = Fs cos θ. From the definition of equipotential, the work done
moving charge along the equipotential surface must be zero. Since neither the displacement nor
the force is zero, the angle between the force and the displacement must be 900. As the
displacement is tangential to the surface, the force, and hence the electric field must be
perpendicular to the surface.
4. A unit disk is defined by the equation 𝑥 2 + 𝑦 2 ≤ 1. Thus the Poisson’s equation is given by
∂2 φ
∂x2
+
∂2 φ
∂y2
ρ
ϵ0
= − 0 , subject to 𝜑 = 0 for 𝑥 2 + 𝑦 2 = 1. Since 0 is constant, the integration are
easily done and we get,
ρ0 2
(𝑥 + 𝑦 2 ) + 𝐶
4ϵ0
ρ
,where C is a constant. The boundary condition on the periphery of the disk gives 𝐶 = + 4ϵ0 .
𝜑(𝑥, 𝑦) = −
0
Thus the solution is
ρ0 2
(𝑥 + 𝑦 2 − 1)
4ϵ0
5. Putting 𝜑(𝑥, 𝑦) = constant gives circles as equipotential contours. The solution above is only
valid within the disk.
𝜑(𝑥, 𝑦) = −
ρ
1 d
r2 dφ
) with
dr
6. Since the potential is spherically symmetric, we have – ϵ = ∇2 𝜑 = 𝑟2 dr (
0
One can easily integrate the relationship to get 𝜌(𝑟) = −
potential has a singularity at r=0 where it has the form 𝜑
φ=
φ0 e−λr
.
r
𝜖0 𝜑0 2 −𝜆𝑟
𝜆 𝑒 . However, note that
𝑟
𝜑
1
= 0. We know that ∇2 ( ) =
𝑟
𝑟
the
−4𝜋𝛿(𝑟). Thus near the origin the charge density is like that of a point charge. Adding the two
1
𝑟
contributions, we have𝜌(𝑟) = 𝜖0 𝜑0 (4𝜋𝛿(𝑟) − 𝜆2 𝑒 −𝜆𝑟 ). The total charge for this distribution
is obtained by integrating this expression over the infinite spherical volume
∞
𝑄 = 4𝜋𝜖0 𝜑0 ∫ 𝛿(𝑟)𝑑3 𝑟 − 𝜖0 𝜑0 𝜆2 4𝜋 ∫ 𝑒 −𝜆𝑟 𝑟𝑑𝑟
0
1
= 4πϵ0 φ0 − 4πϵ0 φ0 λ2 × 2 = 0
λ
(In the second integral above, the factor 4p comes from the angle integration)
7. The equation to the equipotential surface is
2𝑞
√(4−𝑥)2 +𝑦 2 + 𝑧 2
−
3𝑞
√(9−𝑥)2 +𝑦 2 +𝑧 2
= 0. Solving, the
surface is a sphere with its centre at the origin with a radius of 6.
8. The electric field between the plates remain the same. The capacitances reduces by a factor of
two. Since Q has not changed, the potential difference V=Q/C doubles. Thus the stored energy
doubles, the extra energy coming from the work done in pulling the plates apart.
9. The capacitance is given by 𝐶 =
2𝜋𝜖0 𝐿
𝑏
ln( )
𝑎
𝜖 𝐿
= 2𝜋 ln02 = 20 × 10−12 , which gives 𝐿 ≈ 25 cm. Thus
𝑟𝑖𝑛 = 1.25 cm a𝑛𝑑 𝑟𝑜𝑢𝑡 = 2.5 cm . The potential difference between the plates is Q/C = 500 V.
The energy density at a distance r from the axis is found using Gauss’s law. 2𝜋𝑟𝐿 𝐸 = 𝑄𝑒𝑛𝑐𝑙 /𝜖0,
Q
1
which gives, since 𝑄𝑒𝑛𝑐𝑙 = 𝑄 = 10−8 𝐶, 𝐸⃗⃗ =
= 720/r N/C. The energy density is 𝑢(𝑟) =
2πϵ0 L r
1
(2) 𝜖0 𝐸 2 .
Integrating, we get the total energy to be
𝑟𝑜𝑢𝑡
∫
𝑟𝑖𝑛
1
Q2
( ) 𝜖0 E 2 2πrLdr =
ln 2 = 2.5 × 10−6 J
2
4πϵ0 L
1
This agrees with the result one directly agrees by calculating the energy as (2) 𝐶𝑉 2 .
10. The initial charge in the first sphere is 𝑄 = 𝐶𝑉 = 2 × 10−9 𝐶. When the two spheres are
connected, they will have a common potential V. If Q1 and Q2 are the shared charges after the
spheres are connected, we have
𝑄1
𝑉
=
potential to be 20 V. The final energy
𝑄2
, This, together with 𝑄1 + 𝑄2 = 𝑄, gives the common
𝑉
1
( ) (𝐶1 𝑉 2 + 𝐶2 𝑉 2 ) = 2 × 10−8 J, which is one fifth of the
2
−8
initial energy. The lost energy is 8 × 10 𝐽.
Lecture 11
Major correction to text
Replace page 11 by the following : ( Replace the figure also)
Potential due to dielectric
Consider the dielectric to be built up of volume d’. The dipole moment of the volume element is 𝑃′ 𝑑𝜏 ′
. Note that the prime over a coordinate are appropriate to positions within the material volume. The
potential at a point S, whose position vector is 𝑟⃗ is given by
𝑑𝜙 =
1 ⃗⃗⃗⃗
𝑃′ (𝑟⃗⃗⃗⃗′ )(𝑟⃗ − ⃗⃗⃗⃗
𝑟 ′ )𝑑𝜏′
3
4𝜋𝜖0
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
𝑃⃗⃗ (𝑟⃗⃗⃗⃗′ )
⃗⃗⃗⃗
𝑟′
O
⃗⃗⃗
𝑟⃗ − 𝑟′
𝑟⃗
S
Replace page 12
Potential due to the whole volume is
𝜙(𝑟⃗) =
⃗⃗⃗⃗
1
𝑃′ (𝑟⃗⃗⃗⃗′ )(𝑟⃗ − ⃗⃗⃗⃗
𝑟 ′ )𝑑𝜏′
1
1
⃗⃗⃗⃗
∫
=
∫
𝑃′ (𝑟⃗⃗⃗⃗′ ) ∙ ∇′ (
) 𝑑𝜏′
3
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
where we have used
∇′ (
(𝑟⃗ − ⃗⃗⃗⃗
𝑟′)
)=
3
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
1
(Note that the gradient being with respect to primed variable is negative of the gradient with respect to
the unprimed variable, since the function depend on the difference 𝑟⃗ − ⃗⃗⃗⃗
𝑟′ )
Use the vector identity
⃗⃗⃗⃗𝑓) = 𝐴⃗ ⋅ ∇𝑓 + 𝑓 ∇ ⋅ 𝐴⃗
∇ ⋅ (𝐴
⃗⃗⃗) and 𝑓 =
Substitute 𝐴⃗ = 𝑃⃗⃗ (𝑟′
∇′ ⋅ (
𝜙(𝑟⃗) =
1
⃗⃗⃗⃗⃗′ |
|𝑟⃗−𝑟
⃗⃗⃗)
𝑃⃗⃗ (𝑟′
1
1
) = 𝑃⃗⃗ (𝑟⃗⃗⃗⃗′ ) ⋅ ∇′
+
∇′ ⋅ 𝑃⃗⃗ (𝑟⃗⃗⃗⃗′ )
𝑟′|
𝑟 ′ | |𝑟⃗ − ⃗⃗⃗⃗
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
|𝑟⃗ − ⃗⃗⃗⃗
⃗⃗⃗)
1
𝑃⃗⃗ (𝑟′
1
1
⃗⃗⃗) 𝑑𝜏′
∫
∇′ ⋅ (
) 𝑑𝜏 ′ −
∫
∇′ ⋅ 𝑃⃗⃗ (𝑟′
′
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒 |𝑟⃗ − ⃗⃗⃗⃗
𝑟|
𝑟′|
|𝑟⃗ − ⃗⃗⃗⃗
Replace Page 13
The first integral can be converted to a surface integral using divergence theorem, giving
𝜙(𝑟⃗) =
⃗⃗⃗)
⃗⃗⃗)
1
𝜎𝑏 (𝑟′
1
𝜌𝑏 (𝑟′
∫
𝑑𝑆 ′ +
∫
𝑑𝜏′
4𝜋𝜖0 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 |𝑟⃗ − ⃗⃗⃗⃗
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒 |𝑟⃗ − ⃗⃗⃗⃗
𝑟′|
𝑟′|
where the first term is the potential that one would expect for a surface charge density
̂
𝜎𝑏 (𝑟⃗⃗⃗⃗′ ) = 𝑃⃗⃗ (𝑟⃗⃗⃗⃗′ ) ⋅ 𝑛′
̂ being along the outward normal to the surface of volume at the position 𝑟′
⃗⃗⃗ . The second term is the
𝑛′
potential due to a volume charge density
𝜌𝑏 (𝑟⃗⃗⃗⃗′ ) = −∇′ ⋅ 𝑃⃗⃗ (𝑟⃗⃗⃗⃗′ )
The corresponding electric field is given by
𝐸(𝑟⃗) =
⃗⃗⃗)(𝑟⃗ − ⃗⃗⃗⃗
⃗⃗⃗)(𝑟⃗ − ⃗⃗⃗⃗
1
𝜎𝑏 (𝑟′
𝑟′) ′
1
𝜌𝑏 (𝑟′
𝑟′)
∫
𝑑𝑆
+
∫
𝑑𝜏′
3
3
4𝜋𝜖0 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 |𝑟⃗ − ⃗⃗⃗⃗
4𝜋𝜖0 𝑉𝑜𝑙𝑢𝑚𝑒 |𝑟⃗ − ⃗⃗⃗⃗
𝑟′|
𝑟′|