~ from V
E
~
V from E
Electric Potential and E~ Fields
PHYS 272 - David Blasing
Monday June 24th
1 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
E~ Fields are Related to Electric Potentials
If you have the electric potential, you can get the E~ field as follows:
~
Definition: the gradient ∇
~ =(∂ , ∂ , ∂)
∇
∂x ∂y ∂z
The gradient is a vector operator. It operates on a scalar function
and produces a vector function.
2 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
E~ Fields are Related to Electric Potentials
If you have the electric potential, you can get the E~ field as follows:
~
Definition: the gradient ∇
~ =(∂ , ∂ , ∂)
∇
∂x ∂y ∂z
The gradient is a vector operator. It operates on a scalar function
and produces a vector function.
Getting E~ from V:
~ = −( ∂V , ∂V , ∂V )
E~ = −∇V
∂x ∂y ∂z
You ”pass in” a scalar function (like V) from the right and take the
corresponding partial derivative for each component.
2 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
E~ Fields are Related to Electric Potentials
Getting E~ from V:
~ = −( ∂V , ∂V , ∂V )
E~ = −∇V
∂x ∂y ∂z
~ points in direction of maximum increase
Description in words: ∇V
of the electric potential V
~ points in the direction of the maximal
=⇒ So E~ = −∇V
decrease in V.
Now recall that protons feel a force parallel to E~ (maximal
decrease in V). So they are pushed in the direction that decreases
their energy the fastest, which makes sense.
3 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Remember that between the plates of a capacitor, E~ has
magnitude ≈ σ0 and points from the positive to the negative plate
4 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Remember that between the plates of a capacitor, E~ has
magnitude ≈ σ0 and points from the positive to the negative plate
Let x point from the positive plate to the negative plate. Question:
what kind of potential would produce this E~ ?
4 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Question: what kind of potential would produce E~ =
σ
0 x̂?
5 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Question: what kind of potential would produce E~ =
σ
0 x̂?
Now we know that Ex = − ∂V
∂x
5 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Question: what kind of potential would produce E~ =
σ
0 x̂?
Now we know that Ex = − ∂V
∂x
=⇒ V (x) = − σ0 x + V0 with V0 a constant of integration
5 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Question: what kind of potential would produce E~ =
σ
0 x̂?
Now we know that Ex = − ∂V
∂x
=⇒ V (x) = − σ0 x + V0 with V0 a constant of integration
This is just the equation of a line with slope − σ0 , y intercept
of V0
5 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
Question: what kind of potential would produce E~ =
σ
0 x̂?
Now we know that Ex = − ∂V
∂x
=⇒ V (x) = − σ0 x + V0 with V0 a constant of integration
This is just the equation of a line with slope − σ0 , y intercept
of V0
Set V0 to be 0 (we are free here to redefine the reference
energy)
5 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
V (x) = − σ0 x
Let separation distance of the capacitor be s, then v (− 2s ) =
sσ
and v ( 2s ) = −v (− 2s ) = − 2
0
sσ
20
6 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Capacitor Example
V (x) = − σ0 x
Let separation distance of the capacitor be s, then v (− 2s ) =
sσ
and v ( 2s ) = −v (− 2s ) = − 2
0
So V ramps linearly through 0 from
sσ
20
sσ
20
sσ
to − 2
0
6 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Cross Section Deep Inside a Capacitor
Plot of V in dimensionless units
Plate Held at
Positive Electric
Potential
1
In this region:
V ~ −σ x/ (2 ε0)
0.5
Sσ 0
)
V ( 2ǫ
0
−0.5
−1
−10
−8
−6
−4
−2
y position (S)
0
2
4
Plate Held at
Negative Electric
Potential
6
Plate Seperation, S
8
10
2
1.5
1
0.5
0
x position (S)
−0.5
−1
−1.5
7 / 29
~ from V
E
~
V from E
~ from V
E
Capacitor Example
Slope of V and E~ Example
8 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Electric Potentials are Related to E~ Fields
A change in potential can be calculated from the E~ field as follows:
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
9 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Electric Potentials are Related to E~ Fields
A change in potential can be calculated from the E~ field as follows:
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
1
~ = (dx, dy , dz) (in cartesian coordinates)
dl
9 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Electric Potentials are Related to E~ Fields
A change in potential can be calculated from the E~ field as follows:
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
1
~ = (dx, dy , dz) (in cartesian coordinates)
dl
2
R ~rf
is short hand for the x bounds go from xi to xf and
likewise for the y and z bounds
~
ri
9 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Electric Potentials are Related to E~ Fields
A change in potential can be calculated from the E~ field as follows:
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
1
~ = (dx, dy , dz) (in cartesian coordinates)
dl
2
R ~rf
is short hand for the x bounds go from xi to xf and
likewise for the y and z bounds
3
So −
~
ri
R ~rf
~
ri
~ ≡−
E~ · dl
R xf
xi
Ex dx −
R yf
yi
Ey dy −
R zf
zi
Ez dz
9 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Round-Trip Path - Conservation of Energy
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
1
If ~ri = ~rf , then ∆V = 0. This is a statement of conservation
of energy. If you start and end at the same location, then your
energy shouldn’t have changed.
10 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Round-Trip Path - Conservation of Energy
Getting ∆V from an E~ Path Integral
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
1
2
If ~ri = ~rf , then ∆V = 0. This is a statement of conservation
of energy. If you start and end at the same location, then your
energy shouldn’t have changed.
~ is positive making
If you are traveling with E~ then E~ · dl
R ~rf
~ negative. Going against E~ makes ∆V
∆V = − ~r E~ · dl
i
positive.
With E~ , negative ∆V
Against E~ , positive ∆V
10 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
∆V is Path Independent
When E~ is caused by stationary point charges, ∆V = −
is independent of the path.
R ~rf
~
ri
~
E~ · dl
11 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
∆V is Path Independent
When E~ is caused by stationary point charges, ∆V = −
is independent of the path.
R ~rf
~
ri
~
E~ · dl
That is why we could even write the LHS as Vf -Vi . If this was not
true, then ∆V would depend on which particular path we used to
get from ~ri to ~rf
11 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
∆V is Path Independent
When E~ is caused by stationary point charges, ∆V = −
is independent of the path.
R ~rf
~
ri
~
E~ · dl
That is why we could even write the LHS as Vf -Vi . If this was not
true, then ∆V would depend on which particular path we used to
get from ~ri to ~rf
Use this to your advantage, pick whichever path that makes
R ~r
~ the easiest. You will get the same ∆V
evaluating − ~r f E~ · dl
i
11 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples of Paths
Question: which path is easier?
1.
2.
12 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples of Paths
Question: which path is easier?
1.
2.
Answer: path 2 because it moves strictly k with E~ and strictly ⊥
to E~
12 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
~
Examples of dl
~ = dx x̂
1 - dl
~ = −rsin(θ)dθx̂ + rcos(θ)dθŷ
2 - dl
13 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Example: V Inside a Conductor In Static Equilibrium
Remember: in a conductor in equilibrium E~ = ~0
Rf
~ = 0 and hence V = constant
Thus, ∆V = − i E~ · dl
14 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Example: V Inside a Conductor In Static Equilibrium
Remember: in a conductor in equilibrium E~ = ~0
Rf
~ = 0 and hence V = constant
Thus, ∆V = − i E~ · dl
So in a conductor in equilibrium the potential is exactly the
same everywhere inside the conductor. It is a surface of
“equipotential”.
Note: the potential inside a conductor is constant everywhere
but this does not mean it is zero
14 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
15 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
i
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
3
Let the path be always radially in, and label it as along the x
axis, with ri = ∞, rf = r a constant
i
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
3
Let the path be always radially in, and label it as along the x
axis, with ri = ∞, rf = r a constant
4
ri = ∞ =⇒ vi = 0 (charges have no effect infinitely far
away)
i
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
3
Let the path be always radially in, and label it as along the x
axis, with ri = ∞, rf = r a constant
4
ri = ∞ =⇒ vi = 0 (charges have no effect infinitely far
away)
R r 1 ±q
V (r ) = − ∞ ( 4π
2 x̂) · (dx x̂)
0 x
i
5
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
3
Let the path be always radially in, and label it as along the x
axis, with ri = ∞, rf = r a constant
4
ri = ∞ =⇒ vi = 0 (charges have no effect infinitely far
away)
R r 1 ±q
V (r ) = − ∞ ( 4π
2 x̂) · (dx x̂)
0 x
R
r dx
±q
V (r ) = − 4π
2
0 ∞ x
i
5
6
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Examples: V from a Point Charge at a Single Location
V at a single location almost always means referenced to infinity
1
Task derive V from a ±q point charge at a distance r away
2
Start from the general definition:
R ~r
~
∆V = Vf − Vi = − ~r f E~ · dl
3
Let the path be always radially in, and label it as along the x
axis, with ri = ∞, rf = r a constant
4
ri = ∞ =⇒ vi = 0 (charges have no effect infinitely far
away)
R r 1 ±q
V (r ) = − ∞ ( 4π
2 x̂) · (dx x̂)
0 x
R
r dx
±q
V (r ) = − 4π
2
0 ∞ x
i
5
6
7
V (r ) =
±q 1
4π0 r
15 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
V from a Point Charge at a Single Location
V from a Point Charge at a Distance r, Referenced to ∞
V (r ) =
±q 1
4π0 r
16 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
V from a Point Charge at a Single Location
V from a Point Charge at a Distance r, Referenced to ∞
V (r ) =
±q 1
4π0 r
We actually derived this last lecture, here we did it through a
slightly different method
16 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Clicker Question 1
17 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Constant E~ Path Integral
Suppose you have some wonky path but E~ is constant (say E~0 )
along it
18 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
Suppose you have some wonky path but E~ is constant (say E~0 )
along it
∆V simplifies in this case, start with general definition:
∆V = Vf − Vi = −
R ~rf
~
ri
~
E~ · dl
18 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
1
∆V = Vf − Vi = −
R ~rf
~
ri
~
E~ · dl
19 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
1
∆V = Vf − Vi = −
2
=−
R ~rf
~
ri
R ~rf
~
ri
~
E~ · dl
~
E~0 · dl
19 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
1
∆V = Vf − Vi = −
2
=−
3
R ~rf
~
ri
R ~rf
~
ri
~
E~ · dl
~
E~0 · dl
E~0 comes out like any other constant, it isn’t a function of
any spatial coordinate
19 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
1
∆V = Vf − Vi = −
2
=−
3
4
R ~rf
~
ri
R ~rf
~
ri
~
E~ · dl
~
E~0 · dl
E~0 comes out like any other constant, it isn’t a function of
any spatial coordinate
= −E~0 ·
R ~rf
~
ri
~
dl
19 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
1
∆V = Vf − Vi = −
2
=−
3
R ~rf
~
ri
R ~rf
~
ri
~
E~ · dl
~
E~0 · dl
E~0 comes out like any other constant, it isn’t a function of
any spatial coordinate
R ~rf
4
= −E~0 ·
5
Group Question: what is
~
ri
~
dl
R ~rf
~
ri
~
dl?
19 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Constant E~ Path Integral
R ~rf
~
ri
~ is ∆~L, the total change in position.
dl
20 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Constant E~ Path Integral
∆V when E~ is a constant E~0
∆V = −E~0 · ∆~L
21 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Constant E~ Path Integral
∆V when E~ is a constant E~0
∆V = −E~0 · ∆~L
Here ∆V = −|E~0 ||∆~L|cos(θ), θ is the angle between E~ and ∆~L
21 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Constant E~ Path Integral
∆V when E~ is a constant E~0
∆V = −E~0 · ∆~L
Here ∆V = −|E~0 ||∆~L|cos(θ), θ is the angle between E~ and ∆~L
Question: what determines the sign of the change in the electric
potential?
21 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Constant E~ Path Integral
∆V when E~ is a constant E~0
∆V = −E~0 · ∆~L
Here ∆V = −|E~0 ||∆~L|cos(θ), θ is the angle between E~ and ∆~L
Question: what determines the sign of the change in the electric
potential? Just the angle between E~ and ∆~L.
∆V is negative for a path ”parallel” (0o ≤ θ < 90o ) to E~0
∆V is 0 for a path perpendicular (θ = 90o ) to E~0
∆V is positive for a path ”anti-parallel” (90o < θ ≤ 180o ) to E~0
21 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~
1
2
3
The bounds of the integral tell you where you are along the
path, make sure that E~ is correct for that region.
So if E~ is defined piecewise, then −
evaluated piecewise
R ~rf
~
ri
~ must be
E~ · dl
As many times as E~ changes in different regions is how many
integrals you will likely have to evaluate
22 / 29
~ from V
E
~
V from E
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
∆VBA = VB − VA = VB + (VC − VC ) − VA
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
∆VBA = VB − VA = VB + (VC − VC ) − VA
= (VB − VC ) + (VC − VA )
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
∆VBA = VB − VA = VB + (VC − VC ) − VA
= (VB − VC ) + (VC − VA )
= ∆VBC + ∆VCA
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
∆VBA = VB − VA = VB + (VC − VC ) − VA
= (VB − VC ) + (VC − VA )
= ∆VBC + ∆VCA
~ BC ) + (−E~2 · ∆l
~ CA )
= (−E~1 · ∆l
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Piecewise E~ Example
Potential difference across a nonuniform E~ field (two regions, each
with a different but uniform E~ ). Final position is B, and initial is
A. Let the origin be at the left most plate with x measured to the
right.
∆VBA = VB − VA = VB + (VC − VC ) − VA
= (VB − VC ) + (VC − VA )
= ∆VBC + ∆VCA
~ BC ) + (−E~2 · ∆l
~ CA )
= (−E~1 · ∆l
= −E1 ∗ (xC − xA ) + E2 ∗ (xB − xC )
23 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Cavaets
Common error 1: the electric field at a location determines
the potential at that location
24 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Cavaets
Common error 1: the electric field at a location determines
the potential at that location
Common error 2: the electric field at A and B determine ∆V
between those two locations
Rf
~ it is the electric field in the
Since ∆V = − i E~ · dl
intervening region that determines ∆V
24 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Reminder
Read in detail the sections we covered today. There is some
material that I didn’t cover for sake of time, but all of the material
of chapter 17 is fair game for the mid-term Tuesday 07/02 at
8-9:30 pm in PHYS 203.
25 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Clicker Question 2
26 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Clicker Question 3
27 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Clicker Question 4
28 / 29
~ in General
V from E
~ is a constant E~0
∆V when E
Clicker Questions
~ from V
E
~
V from E
Clicker Question 5
29 / 29