1. AP Electrostatics The force between two isolated charges is governed
by Coulomb’s Law:
q1 and q2 are charges
k q1 q2
Fe =
r = distance r2
k = 9 x 109 Nm2/C2
More generally:
εo = permittivity of
free space 1
q1 q2
Fe =
4π εo r2
= x C2/Nm2

2. 1. Use Coulomb’s Law to determine the force between a charge of
+3.5 C and –2.0 C when they are separated by a distance
of 0.50 cm.
q1 = μC
= x 10-6 C
r = cm
= m
q2 = μC
= x 10-6 C 1
q1 q2
εo = x C2/Nm2
Fe =
4π εo r2
(+3.5 x 10-6C)(-2.0 x 10-6C) =
4π (8.85 x C2/Nm2) ( m )2
= N

3. Electric Fields +Q Consider a point charge +Q :
Charge will generate an electric field +q +Q
Field can be “mapped” by the
trajectory of a small charge +q
placed in the region around +Q
Field lines begin on positive charges and end
on negative charges
Field lines never intersect
Strength of the field is determined by the amount of force
on the charge

4. Electric field
intensity F
E = q
2. A charge of 4.5 mC feels a force of 6.5 N when in an
electric field. Find the electric field intensity. F
6.5 N
E = = q C
E = N/C = N/C

5. Electric Field Around a Point Charge
Apply Coulomb’s Law to F +q
E = q +Q
to get 1 Q
E =
4π εo r2
Electric fields are vectors; have a direction

6. 3. Consider a sphere of charge +4.0 μC. Consider the sphere as a
point charge.
(a) Find the electric field intensity 0.60 m away from the sphere. 1 Q
k Q
E = =
4π εo r2 r2
( 9 x 109 Nm2/C2 )( 4.0 x 10-6 C ) =
( 0.60 m )2
E = 1.0 x 105 N/C , away from sphere

7. Electric Potential
Defined as: the work done per unit charge as the charge is
moved in an electric field +q +q W
V = +Q q
Around a Point Charge:
negative, because field pushes
charge away, increasing its KE,
so decreasing its PE 1 Q
V = -
4π εo r
- k Q
V =
OR : r

8. 3. Consider a sphere of charge +4.0 μC. Consider the sphere as a
point charge.
(b) What is the electric potential 0.60 m away from the sphere?
- k Q
V = r
- ( 9 x 109 Nm2/C2 )( 4.0 x 10-6 C ) =
( 0.60 m )
V = Nm/C
Electric potential is a scalar quantity:
no direction, just a magnitude
= J/C
= V

9. 3. Consider a sphere of charge +4.0 μC. Consider the sphere as a
point charge.
(c) If an electron were placed 0.60 m away from the sphere, what
would be its electric potential energy?
Electric potential at that location = V
= J/C
To get energy (in joules), multiply by the charge
So PEe = q V
= ( x C )( J/C )
PEe = 9.6 x J

10. In general, the PE of one charge q1 due to a
second charge q2 : q2 1
PEe = q1 V2
V2 = -
4π εo r
On AP Test Equation Bank, U represents potential energy
Electric
Potential
Energy 1
q1 q2
- k q1 q2
Ue = - =
4π εo r r

11. 4. Consider the system below:
(a) Find the net electric force on charge in the middle.
+ 4.0 μC
- 1.0 μC
+ 1.0 μC
5.625 N
5.625 N q1 q3 q2
8.0 cm
4.0 cm
k q1 q3
(9 x 109 Nm2/C2)(+4.0 x 10-6 C)(-1.0 x 10-6 C)
F13 = = r2
( m )2
= N
to the left
k q2 q3
(9 x 109 Nm2/C2)(-1.0 x 10-6 C)(+1.0 x 10-6 C)
F23 = = r2
( m )2
= N
to the right
Fnet = ( N ) + ( N )
= Fnet = 0

12. 4. Consider the system below:
(b) Find the net electric field at pt. A.
+ 4.0 μC
+ 1.0 μC E2 E1 q1 A q2
8.0 cm
4.0 cm
k q1
(9 x 109 Nm2/C2)(+4.0 x 10-6 C)
E1 = = r2
( m )2
= x 106 N/C
to the right
k q2
(9 x 109 Nm2/C2)(+1.0 x 10-6 C)
E2 = = r2
( m )2
= x 106 N/C
to the left
Enet = ( E6 N ) + ( E6 N )
= Enet = 0

13. 4. Consider the system below:
(c) Find the electric potential at pt. A.
+ 4.0 μC
+ 1.0 μC q1 A q2
8.0 cm
4.0 cm
k q1
(9 x 109 Nm2/C2)(+4.0 x 10-6 C)
V1 = -
= - r
0.080 m
= V
k q2
(9 x 109 Nm2/C2)(+1.0 x 10-6 C)
V2 = -
= - r
0.040 m
= V

14. 4. Consider the system below:
(c) Find the electric potential at pt. A.
+ 4.0 μC
+ 1.0 μC q1 A q2
8.0 cm
4.0 cm
V1 = V V2 = V
Because electric potential is a scalar quantity, just add the individual
potentials to find the net potential:
Vnet = V1 + V2
= ( V ) + ( V )
Vnet = V

15. 5. (a) Determine the net electric field at pt. A.
Electric field at A due to
the +2.0-C charge: E1
k Q1 A
E1 =
r12
15 cm
( 9 x 109 Nm2/C2 )( 2.0 C )
36 cm =
( 0.15 m )2
+ 2.0 C
- 3.0 C
E1 = 8.0 x 1011 N/C , in positive y-direction

16. 5. (a) Determine the net electric field at pt. A.
Electric field at A due to
the -3.0-C charge:
r2 = (15)2 + (36)2 E1
r2 = 1521
r = 39
k Q2 A
E2 = E2
r22 r
15 cm
( 9 x 109 Nm2/C2 )( C )
36 cm =
( 0.39 m )2
+ 2.0 C
- 3.0 C
E2 = x 1011 N/C
or better,
E2 = x 1011 N/C , towards the charge in the direction shown
Net electric field will be the vector sum of the individual fields
Add the vectors by the component method

17. Draw x- and y-components of E2
E1 = 8.0 x 1011 N/C
E2 = x 1011 N/C
Draw x- and y-components of E2
; from geometry, we know small angle in other triangle = θ
; need angle θ E1 15
From SOH-CAH-TOA:
tan θ = = 36
E2x A θ
θ = tan -1 ( )
= 22.6o
E2y E2
15 cm
36 cm θ
+ 2.0 C
- 3.0 C
Then
and
E2x
E2y
cos =
sin =
1.775 x 1011
1.775 x 1011
E2x = x 1011
E2y = x 1010

18. x- and y-components of resultant:
E1x = 0 E1
E1y = 8.0 x 1011 N/C
E2x
E2x = x 1011 N/C A θ
E2y = x 1010 N/C ( downward )
E2y E2
15 cm
36 cm θ
+ 2.0 C
- 3.0 C
x- and y-components of resultant:
Rx = E1x + E2x
= ( 1.64 x 1011 ) = x 1011
Ry = E1y + E2y
= ( 8.0 x 1011 ) + ( x 1010 ) = x 1011

19. So R = 7.5 x 1011 N/C at 77o to the horizontal
Rx = x 1011
Ry = x 1011 E1 R
E2x Ry Ry A
E2y E2
15 cm θ
36 cm Rx
+ 2.0 C
- 3.0 C
7.318 x 1011
R 2 = ( 1.64 x 1011 )2 + ( x 1011 )2
tan θ =
= 4.46
1.64 x 1011
R 2 = x 1023
θ = tan -1 ( 4.46 )
= 77o
R = 7.5 x 1011
So R = 7.5 x 1011 N/C at 77o to the horizontal

20. 5. (b) Determine the electric potential at pt. A.
Potential at A due to
the +2.0-C charge: A
39 cm
15 cm
k Q1
V1 = -
36 cm r1
+ 2.0 C
- 3.0 C
( 9 x 109 Nm2/C2 )( 2.0 C )
= -
( 0.15 m )
V1 = x 1011 V
Potential at A due to the C charge:
k Q2
( 9 x 109 Nm2/C2 )( C )
V2 = -
= - r2
( 0.39 m )
V2 = x 1010 V

21. 5. (b) Determine the electric potential at pt. A.
V1 = x 1011 V
39 cm
15 cm
V2 = x 1010 V
36 cm
+ 2.0 C
- 3.0 C
Because electric potential is a scalar quantity, just add the individual
potentials to find the net potential:
Vnet = V1 + V2
= ( x 1011 V ) + ( x 1010 V )
Vnet = x 1010 V

22. But first, a definition: an equipotential line is a line that
Other general points:
But first, a definition: an equipotential line is a line that
connects places of equal potential in an electric field
( Gravitational analog: the surface of a table is a gravitational
equipotential surface, since every place on the table is the same
height off the floor, and so has the same gravitational PE )
Electric field lines are always perpendicular to equipotential lines
( Again, relate to gravity: to map an electric field, we placed a charge in
the field and released it; to map a gravitational field, place an object
in the field and release it [hold your pencil above the table and release it];
the pencil will hit the table along a line perpendicular to the table; so the
gravitational field line [the path of the pencil] is perpendicular to the
equipotential surface [the table])

23. All static charge on a conductor resides on its surface_ _ _
If charge is placed on a conductor, it can
move around within the conductor _ _
Like charges repel, so they will get as far
from each other as possible (on the surface) _ _ _
The surface of a conductor is an equipotential surface
If charge is placed on a conductor, it can move around on
its surface
; if some of the charge is at a higher potential, it will
move until it is at the same potential as all other charges
( can you think of a gravitational analog?)

24. No electric field can exist on the inside of a conductor
On spherical conductors, charge
distributes itself evenly (density of
charge is equal everywhere)
On non-spherical conductors, charge congregates at points of
high curvature; the higher the curvature, the higher the
charge density

25. A charge +Q is placed inside a spherical conducting shell;
what would the resulting electric field look like?
+Q charge will induce a charge of –Q on the inner surface + + + +Q + + + + +
+Q charge will then be on the outside surface

26. Resulting electric field:+ + + +Q + + + + +

27. Capacitors Devices that store electric charge
Consist of two metal plates that are parallel to each other and
separated by a small distance V
When connected to a battery, charge will flow out of the
battery and onto the plates
Will be stored there indefinitely

28. Two metal plates of area A are separated by a distance d
The capacitance of the capacitor (related to the amount of charge
it can hold) depends on:
- the area A
capacitance increases directly with area
the separation
distance d
the closer the plates are, the greater the
capacitance
the material between
the plates
the presence of a dielectric increases the
capacitance

29. A d
Summarized by:
C = capacitance
K εo A
A = area of plates
C = d
d = separation distance
εo = x C2/Nm2
K = dielectric constant

30. a distance of 0.50 cm. Determine the capacitance.
ex. Two parallel plates of cross-sectional area 5.0 cm2 are separated by
a distance of 0.50 cm. Determine the capacitance.
A = 5.0 cm2
= m2
d = m
Dielectric assumed to be air : K = 1
K εo A
( 1 )( 8.85 x C2/Nm2 )( m2 )
C = = d m
Units fun:
C = x C2/Nm
1 C2 / Nm
= 1 C2 / J
= x farads
= 1 C / J/C
= x F
= 1 C / V
Definition: 1 farad = 1 F = 1 C/V

31. When connected to a battery, a certain amount of charge is stored
on the capacitor V
The amount of charge stored on the capacitor depends on:
- the capacitance C
- the voltage of the battery
Q = charge
Q = C V
C = capacitance
V = voltage

32. ex. If the capacitor in the previous example is placed across a 6-V
battery, find the charge placed on the plates.
C = x F
V = 6.0 V
Q = ?
Q = C V
= ( 8.85 x F )( 6.0 V )
= ( 8.85 x C/V )( 6.0 V )
Q = 5.3 x C

33. Energy Stored in a Charged Capacitor
The energy stored in a charged capacitor can be found by
the expression
Ec = ½ Q V
= ½ C V2
Q = C V
ex. Determine the energy stored in the capacitor in the previous
example.
C = x F
V = 6.0 V
Ec = ½ C V2
= ½ ( 8.85 x F )( 6.0 V )2
Ec = 1.6 x J
Units check:
F V2
= C/V V2
= C V
= C J/C
= J

34. Electric Field in a Charged Capacitor V
Electric field lines begin on positive charges and end
on negative charges
So field lines begin on positive plate and end on negative plate
Away from the edges, field lines are parallel
; at edges field lines arc from positive plate to negative plate
Within the capacitor (away from the edges), field is uniform
throughout the region; has the same intensity near the positive
plate, near the negative plate, or right in the middle

35. Electric Field in a Charged Capacitor d V
The strength of the electric field (electric field intensity)
depends on:
- voltage V of the battery (higher voltage, stronger field)
- separation distance d (if plates are closer together, field is
more intense) V
E = d

36. ex. Determine the electric field intensity inside the capacitor in the
previous examples.
V = 6.0 V d = cm = m V
E = d
6.0 V = m
E = V/m

37. 6. In Millikan’s oil-drop experiment, a drop of oil was suspended
between two parallel plates oriented horizontally. If the drop has a
weight of 2.6 x 10-6 N and it has a charge of –3q, where q is the
charge on an electron, determine the strength of the electric field
between the plates.
q E = Wt.
q q Fe
= q E
Wt.
E =
-3q q
Wt.
2.6 x 10-6 N =
-3( x C )
E = 5.4 x 1012 N/C , downward

38. Capacitors in Series and Parallel
(1) Q = Q1 = Q2 = Q3 C1 - +
(2) V = V1 + V2 + V3 - + - Q + - +
Q1 = C1 V1 C2 V
Q2 = C2 V2 - + - +
Q3 = C3 V3
Q = Ceq V - + - + C3 Q Q1
V1 =
V = C1
Ceq Q Q2 Q3
V2 =
V3 =
Ceq V C2 C3

39. Series: (1) Q = Q1 = Q2 = Q3 C1 (2) V = V1 + V2 + V3 Q Q V = C2 V Ceq- +
(2) V = V1 + V2 + V3 - + - Q + - Q +
V = C2 V
Ceq - + - + Q1 Q2 Q - +
V = + + - + C1 C2 C3 C3 Q Q Q Q + = + = C1 C2 C3
Ceq Q 1 1 1 1
Ceq + + V =
Ceq C1 C2 C3

40. Parallel: Q Q V C1 C2 C3 V Ceq (1) V = V1 = V2 = V3 Q1 = C1 V1 V C1 C2 C3 V
Ceq
- -
(1) V = V1 = V2 = V3
Q1 = C1 V1
(1) Q = Q1 + Q2 + Q3
Q2 = C2 V2
Q = Ceq V
Q3 = C3 V3
Q = C1V1 + C2V2 + C3V3
= C1V + C2V + C3V
= Ceq V
Ceq
= C1 + C2 + C3

41. Kirchoff’s Rules for Circuit Analysis
(1) Current into a junction = Current out of the junction
(2) The sum of the changes in potential around any
closed loop of a circuit = 0
ex. Find the currents I1 , I2 , and I3 .
9.0 V I1
A junction is a place in the circuit where the current splits, or comes together. (see highlighted places) RB RC I2 I3
100 Ω
100 Ω
15 V RA
100 Ω
There are three unknowns ( I1 , I2 , and I3 );
we will need 3 independent
equations to find them;
we will use Kirchoff’s Rules to derive the
3 equations, and then solve them for I1 , I2 , and I3

42. (1) Current into a junction = Current out of the junction
Kirchoff’s Rules:
(1) Current into a junction = Current out of the junction
(2) The sum of the changes in potential around any
closed loop of a circuit = 0
9.0 V I1 I3
Notice that the 9.0-V battery is in series with RC; so the current through the battery is also I3 RB RC I2 I3
100 Ω
100 Ω
15 V RA
100 Ω
Then the current into the red junction is I1 , and the current out
of the junction is the sum of I2 and I3
I1 = I2 + I3

43. (1) Current into a junction = Current out of the junction
Kirchoff’s Rules:
(1) Current into a junction = Current out of the junction
(2) The sum of the changes in potential around any
closed loop of a circuit = 0
(1) I1 = I2 + I3
9.0 V I1
An example of a closed loop
is shown;
there are actually
3 loops to choose from RB RC I2 I3
100 Ω
100 Ω
15 V RA
100 Ω

44. problems it is useful to equate current in a circuit
(1) I1 = I2 + I3
9.0 V I1
When analyzing these
problems it is useful to
equate current in a circuit
with current in a river RB RC + I2 I3
100 Ω
100 Ω
15 V - RA
100 Ω
If the current goes as I2 through RB, it is like the current is flowing
“downhill” (river analogy), and so the potential decreases
But if the current goes through RC as shown, it is like going “against
the current”, or “uphill”, and so the potential increases
For current through a battery: when the current flows from the
positive terminal to the negative (green arrow in diagram), it decreases
potential (uphill to downhill);
when it goes from the negative terminal to the positive (violet arrow above), it is going “uphill”, and
increases potential

45. Consider red loop; start at negative terminal of battery
(1) I1 = I2 + I3
9.0 V I1
Consider red loop; start at
negative terminal of battery
and go around clockwise RB RC + I2 I3
100 Ω
100 Ω
15 V - RA
RA is in series with the 15-V
battery, so the current
through RA is also I1
100 Ω I1
Potential is measured in volts; it will increase or decrease through a
resistor according to Ohm’s Law ( V = I R )
Start at negative terminal:
+ 15
- I2 RB
- I1 RA
= 0
Kirchoff Rule #2
uphill through
battery
goes with current through RB
same for the current through RA
so: I I1 = 0

46. Consider blue loop; start at negative terminal of battery
(1) I1 = I2 + I3
9.0 V I1
(2) I I1 = 0 RB RC + I2 I3
100 Ω
100 Ω
Consider blue loop; start at
negative terminal of battery
and go around
counter-clockwise
15 V - RA
100 Ω I1
Start at negative terminal:
+ 9
- I2 RB
+ I3 RC
= 0
uphill through
battery
goes with current through RB
goes against current through RC,
and so increases potential
so: I I3 = 0
This is the third independent equation, along with (1) and (2) derived
earlier (and listed above); the task is now to solve these equations

47. Note: You must use both of Kirchoff’s Rules
(1) I1 = I2 + I3
Note: You must use both of Kirchoff’s Rules
to get the equations; if you use only one
of the Rules, the equations are not
independent and won’t give correct
answers
(2) I I1 = 0
(3) I I3 = 0
Substitute (1) into (2):
I ( I2 + I3 ) = 0
Use Distributive Property
I I I3 = 0
Combine like terms
I I3 = 0
Add equation (3)
(3) I I3 = 0
I = 0
24 = 300 I2
I2 = A
15 - ( 200 )( ) I3 = 0
I3 = 0
I3 = 0
- 1 = 100 I3
I3 = A

48. Negative sign for current indicates that the current
(1) I1 = I2 + I3
9.0 V I1
(2) I I1 = 0
(3) I I3 = 0 RB RC I2 I3
100 Ω
100 Ω
15 V RA
I2 = A
100 Ω
I3 = A
Negative sign for current indicates that the current
flows in a direction opposite that shown in the
diagram;
that’s okay—don’t change it
Finally, to find I1 :
I1 = I2 + I3
= ( ) + ( )
I1 = A
To be sure of the answers, check
see next slide

49. (1) I1 = I2 + I3
I1 = A
(2) I I1 = 0
I2 = A
(3) I I3 = 0
I3 = A
(1) I1 = I2 + I3 ?
= ( ) =
that was easy
(2) I I1 = 0 ?
( ) ( ) = 0 ?
= 0
so far so good
(3) I I3 = 0 ?
( ) ( ) = 0 ?
( - 1 ) = 0
= 0
yes !!!!

50. ex. Find the current in each resistor.
Choose currents in any
direction you wish; we’ll
choose them like this:
R1 = 3.9 Ω
R3 = 6.7 Ω I1 I3
9.0 V I2
R2 = 1.2 Ω
12 V I1
Note: the current through R4 is
the same as the current
through R1, since they are
in series
R4 = 9.8 Ω
Junction Rule:
(1)
I1 = I2 + I3
Loop Rule:
I I I1 = 0
red loop
(2)
I I2 = 0
I I I1 = 0
green loop
(3)
I I3 = 0

51. (1)
I1 = I2 + I3
(2)
I I2 = 0
(3)
I I3 = 0
Solve these equations by previous methods for I1 , I2 , and I3
OR , use matrix arithmetic and let your TI-84 calculator solve them
Here’s how:
Manipulate the equations so that all of the
unknowns are on one side and the constants
are on the other side; put the unknowns in the
same order
I1 = I2 + I3
I1 - I2 - I3 = 0
I I2 = 0
13.7 I I2 = 12
I I3 = 0
13.7 I I3 = 3

52. I1 - I2 - I3 = 0
1 I I I3 = 0
13.7 I I2 = 12
13.7 I I I3 = 12
13.7 I I3 = 3
13.7 I I I3 = 3
re-write with coefficients
for each unknown
make into
a matrix
1. Find Matrix in your calculator
and enter the values in a 3 x 4
matrix
2. Find the option “rref” and do it
on your matrix
ex. rref ( [A] ) if A is the name
of the matrix
3. Answers will be in the last column
I1 = A
I2 = A
let’s check
I3 = A

53. (1)
I1 - I2 - I3 = 0
I1 = A
I2 = A
(2)
13.7 I I2 = 12
I3 = A
(3)
13.7 I I3 = 3
(1)
I1 - I2 - I3 = 0 ?
(0.722) - (1.75) - (-1.03)
= (0.722) - (1.75) + (1.03)
= 0
I expected that
(2)
13.7 I I2 = 12 ?
13.7 ( ) ( 1.75 ) = 12 ?
= 12
oh my goodness…..
(3)
13.7 I I3 = 3 ?
13.7 ( ) ( ) = 3 ?
( ) =
= 3
sweet

54. Immediately after switch is closed (t = 0) what is the current
RC Circuits
R3 = 3 Ω I3
4 µF
R2 = 12 Ω I2
18 V I1
R1 = 6 Ω
Immediately after switch is closed (t = 0) what is the current
through each resistor?

55. Immediately after switch is closed (t = 0) what is the current
RC Circuits
R3 = 3 Ω I3
4 µF
R2 = 12 Ω I2
18 V I1
R1 = 6 Ω
Immediately after switch is closed (t = 0) what is the current
through each resistor?
Initially, capacitor acts as a short circuit, drawing all current
from R2
I2 = 0

56. I2 = 0 I1 = I3 = 2 A V 18 V I = = = 2 A Req 9 Ω R3 = 3 Ω I3 4 µF I2

57. After the switch has been closed a long time (t = ∞) what is the
18 V I1
R1 = 6 Ω
After the switch has been closed a long time (t = ∞) what is the
current through each resistor?
After the capacitor is charged, it acts like an open circuit

58. I1 = I2 = I3 = 0.86 A V 18 V I = = = 0.86 A Req 21 Ω R3 = 3 Ω I3 4 µF

59. After the switch has been closed a long time (t = ∞) what is the Vc
18 V I1
R1 = 6 Ω
After the switch has been closed a long time (t = ∞) what is the
charge on the capacitor?
Q = C VC

60. I1 = I2 = I3 = 0.86 A VC = I2 R2 = ( 0.86 A )( 12 Ω ) = 10.3 V
4 µF
R2 = 12 Ω I2 Vc
18 V I1
R1 = 6 Ω
R3 = 3 Ω
VC = I2 R2
= ( 0.86 A )( 12 Ω )
R2 = 12 Ω I2 Vc
18 V
= V I1
R1 = 6 Ω

61. After the switch has been closed a long time (t = ∞) what is the Vc
18 V I1
R1 = 6 Ω
After the switch has been closed a long time (t = ∞) what is the
charge on the capacitor?
Q = C VC
= ( 4 µF )( 10.3 V )
= Q = 41 µC