Chapters 15-16: Electrostatics
Law of Charges:
Like charges _______________, opposite charges ____________
In an Atom – what is the charge?
Protons:
Electrons:
Neutrons:
Smallest unit of electric charge = e
SI Unit of charge =
Conversion:
Most atoms are neutral; however, some are ______________; which means that they have gained or lost
____________ and are no longer neutral.
Conductors and Insulators
Conductors:
-IN a conductor, resistivity is __________, which means:
Insulators:
-In an insulator, resistivity is _____________, which means:
Units for Resistivity:
Materials can be charged by contact, which is called ________________
Ex:
A charged conductor brought into contact with a neutral conductor will share charge across the
conductor
**The conductor is always on the outside surface of an object (the inside charge of a conductor is
always ____________). Why?
Ex1: A solid conducting sphere is given a positive charge Q. How is the charge Q distributed in or on
the sphere?
(A) It is concentrated at the center of the sphere.
(B) It is uniformly distributed throughout the sphere.
(C) Its density decreases radially outward from the center.
(D) Its density increases radially outward from the center.
(E) It is uniformly distributed on the surface of the sphere only.
Ex2. A hollow metal sphere of radius R is positively charged. Of the following distances from the
center of the sphere, which location will have the greatest electric field strength?
(A) 0 (center of the sphere) (B) 3R/2 (C) 5R/4 (D) 2R
(E) None of the above because the field is of constant strength
Ex3. A physics problem starts: "A solid sphere has charge distributed uniformly throughout. . . " It
may be correctly concluded that the
(A) electric field is zero everywhere inside the sphere
(B) electric field inside the sphere is the same as the electric field outside
(C) electric potential on the surface of the sphere is not constant
(D) electric potential in the center of the sphere is zero
(E) sphere is not made of metal
Electroscope: charge detector used to detect small electric charges based on:
1. Conduction:
2. Induction:
Figure shows charging by induction:
The only way to prove charges is by repulsion of the leads.
Ex4. Two initially uncharged conductors, 1 and 2, are mounted on insulating stands and are in
contact, as shown above. A negatively charged rod is brought near but does not touch them. With
the rod held in place, conductor 2 is moved to the right by pushing its stand, so that the conductors
are separated. Which of the following is now true of conductor 2?
(A) It is uncharged. (B) It is positively charged. (C) It is negatively charged.
(D) It is charged, but its sign cannot be predicted.
(E) It is at the same potential that it was before the charged rod was brought near.
Coulomb’s Law
- Like charged objects _____________, unlike charges ________________
- Charged objects apply a force on each other called the __________________ force or
____________________ force
- Equation:
o
o
This force is very similar to the ___________________ force, in that it is proportional to
the ____________________________ and the inverse square is proportional to
__________________________________.
Direction: if the charges are attracted, then FE is negative, if the charges are repelled,
then FE is positive.
Biggest difference is that electrostatic force can ________________, gravitational force can not.
Also, k >>G, so at an atomic level, gravitational force is insignificant compared to electrostatic forces
Electric forces obey Newton’s Third Law so: ____________________ = ________________________
When the charges are the same sign, the forces are ______________________
When the charges are opposite signs, the forces are _______________________
If the separate charges touch, the total charge is distributed equally between the charges.
Ex5. Two isolated charges, + q and – 2q, are 2 centimeters apart. If F is the magnitude of the force
acting on charge –2Q, what are the magnitude and direction of the force acting on charge +q?
Magnitude
Direction
(A)
(1/2) F
Toward charge – 2q
(B)
2F
Away from charge –2q
(C)
F
Toward charge – 2q
(D)
F
Away from charge – 2q
(E)
2F
Toward charge – 2q
Ex6. Two identical conducting spheres are charged to +2Q and –Q. respectively, and are separated
by a distance d (much greater than the radii of the spheres) as shown above. The magnitude of the
force of attraction on the left sphere is F1. After the two spheres are made to touch and then are reseparated by distance d, the magnitude of the force on the left sphere is F2. Which of the following
relationships is correct?
(A) 2F1 = F2 (B) F1 = F2 (C) F1 = 2F2 (D) Fl = 4F2 (E) F1 = 8 F2
Superposition: If more than two charges are present, the forces can be added (in one or two
dimensions) to calculate the total force acting on one of the objects.
Ex7. A point charge +q is placed midway between two point charges +3q and –q separated by a
distance 2d. If Coulomb’s constant is k, the magnitude of the force on the charge +q is:
Electric fields are a non-contact force (a field force)
Define:
Electric fields are caused by point charges. If there is a point charge Q, we can detect an electric
field around it by placing a tiny positive test charge (we will call it “q”) at various points in space
near Q and measuring the force the test charge feels, from that we can determine the strength of the
electric field
The electric field strength vector (E) =
From here, we can get an expression for the strength of the electric field created by a point source
of magnitude Q. Equation for electric field at the point charge (calculated exactly the same way that
“g” – the gravitational field strength was calculated in Chapter7):
Units on E:
*please remember, this is a vector which means that we need to determine direction first (the
electric field always travels away from positive and toward negative), then use the equation to find
out the magnitude of the electric field. We are dividing by q only because the charge of it doesn’t
matter, since its purpose was to test the electric field at Q.
Visualizing the Electric Field
______________ ___________________ help you visualize the electric field. They:
Drawing field lines:
1.
2.
3.
4.
5.
Ex8. The electric field E just outside the surface of a charged conductor is
(A) directed perpendicular to the surface (B) directed parallel to the surface
(C) independent of the surface charge density (D) zero (E) infinite
Ex9. The diagram above shows electric field lines in an isolated region of space containing two
small charged spheres, Y and Z. Which of the following statements is true?
(A) The charge on Y is negative and the charge on Z is positive.
(B) The strength of the electric field is the same everywhere.
(C) The electric field is strongest midway between Y and Z.
(D) A small negatively charged object placed at point X would tend to move toward the right.
(E) Both charged spheres Y and Z carry charge of the same sign.
Ex10. The figure above shows two particles, each with a charge of +Q, that are located at the
opposite corners of a square of side d. What is the direction of the net electric field at point P ?
Ex11. A positive charge of 10–6 coulomb is placed on an insulated solid conducting sphere. Which of
the following is true?
(A) The charge resides uniformly throughout the sphere.
(B) The electric field inside the sphere is constant in magnitude, but not zero.
(C) The electric field in the region surrounding the sphere increases with increasing distance from
the sphere.
(D) An insulated metal object acquires a net positive charge when brought near to, but not in
contact with, the sphere.
(E) When a second conducting sphere is connected by a conducting wire to the first sphere, charge
is transferred until the electric potentials of the two spheres are equal.
Ex12. An electron is accelerated from rest for a time of 10–9 second by a uniform electric field that
exerts a force of 8.0 x 10–15 Newton on the electron. What is the magnitude of the electric field?
(A) 8.0 × 10–24 N/C (B) 9.1 × 10–22 N/C (C) 8.0 × 10–6 N/C (D) 2.0 × 10–5 N/C (E) 5.0 × 104 N/C
Ex13. A point P is 0.50 meter from a point charge of 5.0 × 10–8 coulomb. The intensity of the electric
field at point P is most nearly
(A) 2.5 × 10–8 N/C (B) 2.5 × 101 N/C (C) 9.0 × 102 N/C (D) 1.8 × 103 N/C (E) 7.5 × 108 N/C
Electricity
Force:
Gravity
Field Strength:
Field Strength:
Constant:
Charge Units:
Mass units:
Electric Potential Energy
We have already learned about three types of Energy:
Now, we are adding in another, electrical potential energy.
In the same way that electric force is related to gravitational force; electric potential energy is
related to gravitational potential energy…
Gravitational Potential Energy assumes a constant gravitational field, and the height that something
is above the baseline in that constant gravitational field is always ____________________________________.
Electric potential energy in a constant electric field has a displacement that is always
_______________________________________. (Draw this out)
The change in the Electric Potential Energy in a constant electrical field gives us the equation:
**This equation is ONLY for a charge in a constant electric field…
*When ΔPE is negative, then Kinetic energy is increasing, when ΔPE is positive, the Kinetic energy is
decreasing…
The electric field itself is constant, however, a particle has different potential energy at different
points in the field. This difference is called the electric potential difference. The electric potential
difference across the electric field or __________________________ can also be computed.
**if we are comparing this to gravity, this would be the “g*h” part of PE=mgh…a charge travels
through a change in _______________________ like a falling mass experiences a change in height…
**This is a scalar property
The graph of Work vs charge gives a slope of the electric potential difference:
Ex14. Two large parallel conducting plates P and Q are connected to a battery of emf E, as shown
above. A test charge is placed successively at points I, II, and III. If edge effects are negligible, the
force on the charge when it is at point III is
(A) of equal magnitude and in the same direction as the force on the charge when it is at point I
(B) of equal magnitude and in the same direction as the force on the charge when it is at point II
(C) equal in magnitude to the force on the charge when it is at point I, but in the opposite direction
(D) much greater in magnitude than the force on the charge when it is at point II, but in the same
direction
(E) much less in magnitude than the force on the charge when it is at point II, but in the same
direction
Two large, flat, parallel, conducting plates are 0.04 m apart, as shown above. The lower plate is at a
potential of 2 V with respect to ground. The upper plate is at a potential of 10 V with respect to
ground. Point P is located 0.01 m above the lower plate.
Ex15. The electric potential at point P is
(A) 10 V (B) 8 V (C) 6 V (D) 4 V (E) 2 V
Ex16. The magnitude of the electric field at point P is
(A) 800 V/m (B) 600 V/m (C) 400 V/m (D) 200 V/m (E) 100 V/m
Ex17. Consider the two oppositely charged plates as shown in the diagram. At which of the marked
points shown in the diagram would a positively charged particle have the greatest electrical
potential energy?
(A) A (B) B (C) C (D) D (E) E
Ex18. How much work would be required to move a 4 coulomb charge 6 meters parallel to a 24 N/C
electric field?
(A) 0 J (B) 24 J (C) 96 J (D) 144 J (E) 576 J
Above, we learned about how the electric potential difference (voltage) changed with the distance
travelled in a constant electric field…now lets talk about how to determine voltage from a point
charge…
Point charges also have electric fields, and in order to draw them, we need to look back at the
previous electric fields that we drew: What is the voltage of points along the same horizontal axis?
*these lines are called _________________________________ ____________________________ and are the key to
addressing the voltage around a point charge…
Equipotential Lines:
In the same way that a topographical map shows you areas that are the same altitude (or, in terms
of physics, equal gravitational potential), lines connecting points of equal electrical potential are
called equipotential lines.
Characteristics of these lines:
1.
2.
3.
Voltage, unlike Electric Field, is NOT a vector! So if you have MORE than one charge you don’t need
to use vectors. Simply add up all the voltages that each charge contributes since voltage is a
SCALAR.
Equation for Voltage around a point charge:
Units:
This is a _________ quantity, so a lot easier to calculate than the electric field strength. Simply sum up
the charges acting on a point charge (don’t forget your signs), and you can determine voltage with
the given equation.
Since we know the relationship between Voltage and change in potential energy, you can also
calculate the potential energy of a point charge:
A sphere of radius R has positive charge Q uniformly distributed on its surface
Ex19. Which of the following represents the magnitude of the electric field E and the potential V as
functions of r, the distance from the center of the sphere, when r < R ?
E
V
(A)
0
kQ/R
(B)
0
kQ/r
(C)
0
0
(D)
kQ/r2
0
(E)
kQ/R
0
Ex20. Which of the following represents the magnitude, of the electric field E and the potential V as
functions of r, the distance from the center of sphere, when r > R ?
E
V
(A) kQ/R2
kQ/R
(B) kQ/R
kQ/R
(C) kQ/R
kQ/r
(D) kQ/r2
kQ/r
(E) kQ/r
kQ/r2
Ex21. A particle of charge Q and mass m is accelerated from rest through a potential difference V,
attaining a kinetic energy K. What is the kinetic energy of a particle of charge 2Q and mass m/2 that
is accelerated from rest through the same potential difference?
(A) ¼ K (B) ½ K (C)K (D) 2K (E) 4K
Ex22. The square of side a above contains a positive point charge +Q fixed at the lower left corner
and negative point charges –Q fixed at the other three corners of the square. Point P is located at the
center of the square.
a. On the diagram, indicate with an arrow the direction of the net electric field at point P.
b. Derive expressions for each of the following in terms of the given quantities and fundamental
constants.
`i. The magnitude of the electric field at point P
`ii. The electric potential at point P
c. A positive charge is placed at point P. It is then moved from point P to point R, which is at the
midpoint of the bottom side of the square. As the charge is moved, is the work done on it by the
electric field positive, negative, or zero?
____Positive ____Negative ____Zero
Explain your reasoning.
d. i: Describe one way to replace a single charge in this configuration that would make the electric
field at the center of the square equal to zero. Justify your answer.
ii. Describe one way to replace a single charge in this configuration such that the electric potential
at the center of the square is zero but the electric field is not zero. Justify your answer.
We have spoken a lot about charged parallel plates with a constant electric field between them, and
we can actually do this in real life…
Often used in physics, two parallel plates of opposite charge, a fixed distance away from each other,
create a ___________________________________ – an electrical device used to store charge
Electrical Field between the two plates, except at the edges, is _______________.
Draw one below, be sure to include the direction of charge flow:
Capacitor:
The parallel plates are separated from each other by some sort of insulator. Examples of insulators
are:
Once the parallel plates are charged, they are disconnected and the charge is stuck to the plates
until they are reconnected…a real life example of this is:
The amount of charge a capacitor can store for a given potential difference across the plates =
_____________________________ _______________________________
Abbreviation:
Equation:
Units:
**If the charge is isolated, then it stays constant in the capacitor…
A Capacitor is a ________________ _____________________ _____________________
The charges are located on opposite plates of a capacitor, which causes them to exert forces on each
other, and allows them to store energy.
Designing Capacitors
Three things determine the amount of charge a capacitor can store:
1.
2.
3.
Equation:
*ε (epsilon) is determined by the material between the plates
Base line ε:
**only for if your material is in a vacuum or air gap, if something other than air:
(a material with a higher dielectric constant increases the capacitance)
What is the purpose of a dielectric?
The energy stored in a capacitor.
Equation:
Units:
Ex23. An isolated capacitor with air between its plates has a potential difference Vo and a charge
Qo. After the space between the plates is filled with oil, the difference in potential is V and the
charge is Q. Which of the following pairs of relationships is correct?
(A)Q=Qo and V > Vo (B) Q = Qo and V < Vo (C) Q > Qo and V = Vo (D) Q < Qo and V < Vo
(E) Q > Qo and V > Vo
Ex24. Two square parallel–plate capacitors of capacitances C1 and C2 have the dimensions shown
in the diagrams above. The ratio of C1 to C2 is
(A) 1 to 4 (B) 1 to 2 (C) 1 to 1 (D) 2 to 1 (E) 4 to 1
Ex25. A parallel–plate capacitor has a capacitance Co. A second parallel–plate capacitor has plates
with twice the area and twice the separation. The capacitance of the second capacitor is most nearly
(A) ¼C0 (B) ½C0 (C) C0 (D) 2 C0 (E) 4 C0
Ex26. A 20 μF parallel–plate capacitor is fully charged to 30 V. The energy stored in the capacitor is
most nearly
(A) 9 × 103 J (B) 9 × 10–3 J (C) 6 × 10–4 J (D) 2 × 10–4 J (E) 2 × 10–7 J
Ex27. A potential difference V is maintained between two large, parallel conducting plates. An
electron starts from rest on the surface of one plate and accelerates toward the other. Its speed as it
reaches the second plate is proportional to
(A) 1/V
(B)1/√V
(C) √V (D) V (E) V2
Ex28. Two parallel conducting plates, each of area 0.30 m2, are separated by a distance of 2.0 × 10–2
m of air. One plate has charge +Q; the other has charge –Q. An electric field of 5000 N/C is directed
to the left in the space between the plates, as shown in the diagram above.
a. Indicate on the diagram which plate is positive (+) and which is negative (–).
b. Determine the potential difference between the plates.
c. Determine the capacitance of this arrangement of plates.
An electron is initially located at a point midway between the plates.
d. Determine the magnitude of the electrostatic force on the electron at this location and state its
direction.
e. If the electron is released from rest at this location midway between the plates, determine its
speed just before striking one of the plates. Assume that gravitational effects are negligible.
Name
Symbol
Units
Equation