1. k = 4.0 E = 2.82 x 107 V/m C = 1 x 10-6 = V = 100 mV = 0.1 V V = E d1.
V = E d
k = 4.0
0.1 = E (3.54 x 10-9)
C= 1μF = 1 x 10-6 F
A = 1 cm2
E = 2.82 x 107 V/m 2
1 cm2
1 m =
1 x 10-4 m2 x
100 cm A
C =
keo d
1 x 10-4
1 x 10-6 =
(4.0)(8.85 x 10-12) d
d = 3.54 nm

2. k = 1.0 ; 5.0 C = 0.04 = PE = ½ QV C = C = 0.20 μF C= 0.04 μF Q Q V2. Q Q
k = 1.0 ; 5.0
C =
0.04 = V
200
V = 200 V
Q = 8μC
PE = ½ QV
(a)
PE = ½ (8)(200)
PE = 800 μJ
(b)
all this stays the same for the capacitor A
keo
C = d
this is the only thing that changes about the capacitor
Simply multiply the original capacitance by 5
C = 0.20 μF

3. C = 0.04 = C = 0.2 = PE = ½QV Q Q Q 8 V 200 V V Q = 8μC V = 40 V
Since the battery is disconnected, the charge on the capacitor must remain the same (there is no place for it to go)
(c)
original
new Q Q
C =
0.04 = Q 8 V
C =
0.2 =
200 V V
Q = 8μC
V = 40 V
NOTE: Had the battery remained connected, the voltage on the capacitor would have remained the same (thus the charge would have been different)
PE = ½QV
PEi + Wc = PEf
PE = 160 μJ
800 + Wc = 160
PE = ½ (8)(40)
Wc = μJ

4. C= 300 pF
original 3. Q Q
k = 4.8
C =
300 x = V
120
V = 120 V
Q = 3.6 x 10-8 C
all this stays the same for the capacitor A
keo
C = d
this is the only thing that changes about the capacitor
Simply multiply the original capacitance by 4.8
C = 1440 pF
new Q
1440 x =
Q = 1.73 x 10-7 C Q
120
C = V
Q = 1.36 x 10-7 C from the battery
Change in Q = Qf - Qi
1.73 x x 10-8 =

5. C1 = 4 = C2 = 4 = 4 = C3 = 2 = Cs = Q1 Q1 V1 12 Q2 24 V2 V2 24 Q3 V34. Q1 Q1
C1 =
4 = V1 12 C2 C3 Q2 24
C2 =
4 = V2 V2
12 V 24 Q3
4 =
C3 = V3 V3
Voltages are the same in
Parallel
both groups get the same voltage V1 = 12 V
Charge is the same in
Series
Cs = 2 μF
Q2 = Q3 Qs Qs
2 =
Cs =
Qs = 24 μC Vs 12

6. Put what we know into a table
capacitor
C (μF)
Q (μC)
V (V)
PE (J) C1 C2 C3 4 48 12
2.88 x 10-4
7.2 x 10-5 4 24 6 4 24 6
7.2 x 10-5
use PE = ½QV to determine the PE

7. 5.
Determine the charge stored on the 40 μF capacitor when connected to the 16 V battery. Q Q
Q = 640 μC
C =
40 = V 16
Since the battery is disconnected, the charge must remain the same
This is then the charge on both capacitors when connected
Qp = 640 μC
Cp = C1 + C2
When the two capacitors
are connected, they form
a parallel group
Cp = = 70 μF

8. 70 = Cp = 40 = C40 = 30 = C30 = 640 Qp Vp = 9.14 V Vp Vp Q Q40
This is then the voltage
across each capacitor Q
Q40
Q40 = 366 μC
40 =
C40 =
9.14 V Q
Q30
Q40 = 274 μC
30 =
C30 =
9.14 V
640 μC

9. TRY # 6 capacitor C (μF) Q (μC) V (V) PE (J) C1 C2 C3 16 267 16⅔
2.2 x 10-3
1.1 x 10-3 12
160
13⅓ 8
107
7.1 x 10-4
13⅓