1. Electrostatics Review Examples
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
q1 q2 1
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

2. 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

3. 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. Q
k Q 1
k = 9 x 109 Nm2/C2
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

4. 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

5. 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

6. 4. (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

7. 4. (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

8. 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

9. 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

10. 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 R
+ 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

11. 4. (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

12. 4. (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 = 5.1 x 1010 V

13. two parallel plates oriented horizontally. If the drop has a weight of
5. 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 a electron,
determine the strength of the electric field between the plates.
If a negative charge is placed in an electric
field as at right, there will be a force on
the charge attracting it upward qE E
-3q
Fe = q E , where E = electric
field intensity
m g
To be suspended, or “float” in the field, this upward force will match
the weight of the object
Fe = Wt.
q E = m g

14. two parallel plates oriented horizontally. If the drop has a weight of
5. 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 a electron,
determine the strength of the electric field between the plates.
q E = m g qE
q q E
m g
-3q
E = q
m g
2.6 x 10-6 N =
3 ( x C )
E = 5.4 x 1012 N/C

15. 6. An electron is moved through a potential difference of 500 volts.
(a) Determine the amount of work done on the charge. W q
Potential
Difference q
V = q
W = q V
qe = x C
= ( x C )( 500 V )
= ( x C )( 500 J/C )
W = x J

16. 6. An electron is moved through a potential difference of 500 volts.
(b) Assuming all of the work done went into kinetic energy and the charge
started from rest, find the final speed of the charge.
W = x J
KE = x J
2 (
KE = ½ m v2
) 2
2 KE = m v2 m m
2 KE
v2 =
me = x kg m
2 KE
2 ( 8.01 x J )
v = = m
9.11 x kg
v = 1.3 x 107 m/s

17. 7. Two parallel plates of cross-sectional area 5
7. Two parallel plates of cross-sectional area 5.0 cm2 are separated by a
distance of 0.50 cm. Determine the capacitance of the plates in farads.
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 check:
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

18. 8. If the capacitor in #7 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

19. 9. Determine the electric potential energy of the capacitor in #7.
C = x F
V = 6.0 V
Q = 5.3 x C
Uc = ½ C V2
= ½ ( 8.85 x F)( 6.0 V )2
= ½ ( 8.85 x C/V)( 6.0 V )2
= 1.6 x C/V
= 1.6 x C/ J/C
Uc = 1.6 x J

20. 10. Determine the electric field intensity inside the capacitor in #7.
V = 6.0 V
d = m V
Electric Field
in a Capacitor
E = d
6.0 V = m
E = V/m

21. 11. Three capacitors are hooked in series across a 6.0-V battery. The
capacitors have capacitances of 3.0 F, 2.0 F, and 1.0 F, respectively. Find
(a) the equivalent capacitance of the circuit.
3 F
6 V
2 F
6 V
Ceq
1 F
For capacitors in series, 1 1 1 1 1 1 1 = + + = + +
Ceq C1 C2 C3 3 2 1 2 3 6 11 1 = + + = = 6 6 6 6
Ceq 6
Ceq =
= Ceq = F 11

22. 11. Three capacitors are hooked in series across a 6.0-V battery. The
capacitors have capacitances of 3.0 F, 2.0 F, and 1.0 F, respectively. Find
(b) the total amount of charge on the capacitors
3 F - + - + + +
6 V
2 F
6 V
Ceq = F - - - + - +
1 F
Qtotal = Ceq V
= ( 0.54 F )( 6.0 V )
= ( 0.54 C/V )( 6.0 V )
Qtotal = 3.3 C
For capacitors in series, Qtotal = Q1 = Q2 = Q3
Negative charge comes out of battery to the plate of the 1-F capacitor; which repels
an equal amount of negative charge to the 2-F capacitor; which repels an equal
amount of charge to the 3-F capacitor; which repels an equal amount to the battery;
charge on each capacitor must be the same

23. Find (a) the equivalent capacitance of the circuit.
12. The capacitors in #11 are now hooked in parallel across the 6.0-V battery.
Find (a) the equivalent capacitance of the circuit.
6 V
3 F
2 F
1 F
6 V
Ceq
For capacitors in parallel,
Ceq = C1 + C2 + C3 =
= Ceq = 6 F

24. 12. The capacitors in #11 are now hooked in parallel across the 6
12. The capacitors in #11 are now hooked in parallel across the 6.0-V battery.
Find (a) the total amount of charge on the capacitors + + + + + +
6 V
3 F
2 F
1 F
6 V
Ceq = 6 F - - - - - -
Qtotal = Ceq V
= ( 6.0 F )( 6.0 V )
= Qtotal = 36 C
For capacitors in parallel, Qtotal = Q1 + Q2 + Q3
Q1 = C1 V
= ( 3.0 F )( 6.0 V ) = 18 C
Q2 = C2 V
= ( 2.0 F )( 6.0 V ) = 12 C
36 C
Q3 = C3 V
= ( 1.0 F )( 6.0 V ) = 6 C
Negative charge comes out of battery to the plate of the 3-F capacitor; additional
negative charge goes to the 2-F capacitor; and additional charge to the 1-F capacitor;
total amount of charge that comes out of the battery is sum of the charge on each
individual capacitor

25. 13. Determine the electric potential energy of the charged capacitors in #12.
6 V
3 F
2 F
1 F
6 V
Ceq = 6 F
Q1 = 18 C
Q2 = 12 C
Q3 = 6 V
Uc = ½ C V2
U1 = ½ C1 V2
= ½ ( 3.0 F )( 6.0 V )2
= U1 = 54 J
U2 = ½ C2 V2
= ½ ( 2.0 F )( 6.0 V )2
= U2 = 36 J
U3 = ½ C3 V2
= ½ ( 1.0 F )( 6.0 V )2
= U3 = 18 J

26. 14. An oil drop that has a net charge of 3e- is suspended between two
parallel plates 1.0 cm apart and hooked across a potential difference
of 3.0 V. Determine the weight of the oil drop.
If a negative charge is placed in an electric
field as at right, there will be a force on
the charge attracting it upward qE E
Fe = q E , where E = electric
field intensity
3e-
m g
Electric Field can
be found by V
E = d
3.0 V =
= V/m m

27. 14. An oil drop that has a net charge of 3e- is suspended between two
parallel plates 1.0 cm apart and hooked across a potential difference
of 3.0 V. Determine the weight of the oil drop.
Fe = q E
E = V/m qE
To be suspended, this upward force will match
the weight of the drop E
3e-
Wt. = Fe
= q E
m g
= 3 ( x C )( 3000 V/m )
Wt. = 1.4 x N
For fun, prove to yourself that C V/m = N