PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
CAPACITANCE
Question 31
How much charge flows from a 12.0 V battery when it is connected to a 2.00 µF capacitor?
q = CV = 2.00 × 10−6 × 12.0 = 24.0 µC
Question 32
The two plates of a capacitor carry +1500 µC and -1500 µC of charge respectively when the
potential difference is 300 V. Calculate the capacitance.
Charge on the capacitor is 1500 µC. Therefore
Q
V
1500 × 10−6
=
300
= 5.00 × 10−6 F
C =
Question 33
Calculate the magnitude of the electric field between the plates of a 20.0 µF capacitor if they
are 2.00 mm apart and each has a charge of 300 µC?
The electric field between two uniformly charged parallel plates (ie a capacitor) is given
by
V
E=
d
where V is the potential difference between the plates and d is the space between them.
Therefore
Q
V
=
d
dC
300 × 10−6
=
(2 × 10−3 )(20.0 × 10−6 )
= 7.50 × 103 V m−1 (N C−1 )
E =
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 34
Three capacitors having capacitances of 0.160 µF, 0.220 µF and 0.470 µF are connected in
parallel and charged to a potential difference of 240 V.
(a) Determine the charge on each capacitor.
(b) What is the total capacitance of the combination?
(c) What is the total charge acquired?
(a) The charge on a capacitor is give by q = CV , hence
q1 = (0.16 × 10−6 ) × 240 = 3.84 × 10−5 C
q2 = (0.22 × 10−6 ) × 240 = 5.28 × 10−5 C
iq3 = (0.47 × 10−6 ) × 240 = 11.3 × 10−5 C
(b) For parallel capacitors, the equivalent capacitance is CEQ = C1 + C2 + C3 + . . .
∴ CEQ = (0.16 + 0.22 + 0.47) × 10−6 = 0.850 µF
(c) The total charge is, from part (a)
Q = q1 + q2 + q3 = (3.84 + 5.28 + 11.3) × 10−5 = 20.4 × 10−5 C
and from part (b)
Q = CEQ V = 0.85 × 10−5 × 240 = 20.4 × 10−5 C
Question 35
A 6.00 µF and a 4.00 µF capacitor are connected in series to a 60.0 V battery.
(a) Calculate the equivalent capacitance.
(b) What is the charge on each capacitor?
(c) Determine the voltage accross each capacitor.
(a) For capacitors in series, the equivalent capacitance is
1
1
1
=
+
CEQ
C1 C2
C1 C2
∴ CEQ =
C1 + C2
6×4
=
6+4
24
=
10
= 2.40 µF
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(b) For capacitors in series, the charge on each capacitor is the same, therefore
Q = CEQ × V
= (2.40 × 10−6 ) × (60.0)
= 144 µC
(c) The voltage on each capacitor is then
Q
144 × 10−6
=
C1
6 × 10−6
= 24.0 V
144 × 10−6
Q
=
=
C2
4 × 10−6
= 36.0 V
V1 =
V2
Question 36
Two identical capacitors are connected in series and found to have charge of magnitude
200 nC on each plate when the combination is connected to a power supply of 50.0 V. Find
the capacitance of each capacitor.
For capacitors in series
C1 C2
C1 + C2
C2
=
2C
1
C
=
2
since C1 = C2 . The charge on capacitors in series is the same, therefore
CEQ =
1
Q = CEQ V = CV
2
2Q
∴C =
V
2 × 200 × 10−9
=
50
= 8.00 nF
Question 37
A 0.10 nF capacitor is charged to 50 V. The charging battery is then disconnected and replaced by a second capacitor. If the measured potential difference drops to 35 V, what is the
capacitance of the second capacitor?
The charge on the first capacitor is
q1 = C1 V = 0.1 × 10−9 × 50 = 5.0 nC
page 3 of 8
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
When the battery is replaced with a capacitor C, charge is conserved and the measured
voltage is 35 V, which is the same across both capacitors. The charge on q1 is now
q1 = 0.1 × 10−9 × 35 = 3.5 nC
Therefore the charge on C is now 5.0 − 3.5 = 1.5 nF, hence the capacitance C is
C=
q
1.5 × 10−9
=
= 4.3 × 10−11 = 43 pF
V
35
Question 38
Calculate the charge on each capacitor shown below
(a) when switch S is open and
(b) when switch S is closed.
(c) Determine also the change in energy stored in the capacitor A when S is closed.
(a) When switch S is open, capacitors A and B are in series with each other, so the
equivalent capacitance is
page 4 of 8
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
12 × 4
CA CB
= 3.00 µF
=
CA + CB
12 + 4
Similarly, capacitors C and D are in series, giving the same equivalent capacitance.
Thus the total charge is
(AB)
CEQ
(AB)
=
(CD)
Q = CV = (CEQ + CEQ )V = (3 + 3) × 10−6 × 240 = 1.44 × 10−3 C
Since each series combination has the same capacitance, the total charge is shared
equally between them
(AB)
CEQ
(AB)
= CEQ
=
1.44 × 10−3
= 720 µC
2
(Or, finding the charge on each series combination individually:
Q = CEQ × V = (3 × 10−6 ) × 240 = 720 µC
(AB)
(CD)
whic is the same for CEQ .)
Now, since for capacitors in series the magnitude of the charge must be the same on
each capacitor
(AB)
(CD)
qA = qB = CEQ = CEQ = qC = qD = 720 µC
(b) When switch S is closed, both parallel combinations has the same capacitance:
(AC)
(BD)
CEQ = CA + CC = CB + CD = CEQ
(AC)
Now CEQ
(BD)
and CEQ
= 16.0 µF
are in series with one another so the total capacitance C is
(AC)
C=
(BD)
CEQ CEQ
(AC)
(BD)
CEQ + CEQ
=
page 5 of 8
162
= 8.00 µF
2 × 16
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
The total charge Q is given by
Q = CV = (8 × 10−6 ) × 240 = 1.92 × 10−3 C
which is the charge on each parallel combination (since these are in series and the
magnitude of the charge must be the same for each capacitor). The voltage drop
across each parallel combination is
V (AC) =
Q
(AC)
CEQ
=
Q
(BD)
CEQ
=
1.92 × 10−3
= 120 V
16 × 10−6
The charge on each capacitor can be found using the voltage drop across it:
qA = CA VA = (12 × 10−6 ) × 120 = 1.44 × 10−3 C
qC = (4 × 10−6 ) × 120 = 4.80 × 10−4 C
and it can be seen that qA = qD and that qC = qB .
(c) The change in energy stored ∆W in the capacitor A when S is closed can be found by
∆W = |WOPEN − WCLOSED |
2 2 qA
q
A
= −
2CA OPEN
2CA CLOSED (720 × 10−3 )2 (1.44 × 10−3 )2 = −
2 × 12 × 10−6
2 × 12 × 10−6 = 6.48 × 10−2 J
Question 39
In the circuit shown, assume that C1 = 10.0 µF, C2 = 5.00 µF, C3 = 4.00 µF and V0 = 100 V.
(a) Find the equivalent capacitance of the combination.
(b) Suppose the capacitor C3 breaks down electrically becoming equivalent to a zeroresistance path. What changes to the charge and potential difference occur for capacitor C1 ?
(a) For the parallel combination, the equivalent capacitance is
page 6 of 8
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
C|| = C1 + C2 = 10 + 5 = 15.0 µF
The total equivalent capacitance CEQ is then
CEQ =
C|| C3
15 × 4
= 3.16 µF
=
C|| + C3
15 + 4
(b) To find the charge q1 , one needs to first find the voltage drop V|| across the parallel
combination. First, find the total charge Q:
Q = CEQ V = 3.16 × 10−6 × 100 = 316 µC
This charge is the charge for the parallel combination and the charge q3 on C3 , since
they are in series. The voltage drop across capacitor C3 is then
q3
316 × 10−6
=
= 78.9 V
C3
4
The voltage drop across the parallel combination must be
V3 =
V|| = V0 − V3 = 100 − 78.9 = 21.1 V
The voltage drop across the combination V|| is the same for both C1 and C2 , therefore
the charge q1 for the capacitor C1 is
q1 = C1 V|| = (10 × 10−6 ) × 21.1 = 211 µC
When the capacitor C3 is shorted (becoming equivalent to a zero-resistance path), the
voltage across it is zero (V3 = 0) and so V|| = V0 = 100 V. The charge q1 on capacitor
C1 is now
q1 = C1 V|| = (10 × 10−6 ) × 100 = 1.00 × 103 µC
Question 40
Calculate the energy stored in a 600 pF capacitor that is charged to 100 V.
The energy stored is
q2
1
W = qV =
2
2C
1
CV 2
2
1
=
(600 × 10−12 )(1002 )
2
= 300 × 10−8
= 3.00 × 10−6 J
=
page 7 of 8
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 41
It takes 6.00 J of energy to move a 2000 µC charge from one plate of a 5.00 µF capacitor to
the other. Calculate the charge on each plate.
The work done is given by
W = qV ∴ V =
W
6
=
= 3.00 kV
q
2000 × 10−6
The charge on each plate of the capacitor is then
Q = CV = (5 × 10−6 ) × (3 × 103 ) = 15 × 10−3 = 1.50 × 10−2 C
Question 42
A 16.0 µF and a 4.00 µF capacitor are connected in parallel and charged by a 22.0 V battery.
What voltage is required to charge a series combination of the two capacitors with the same
total energy?
||
The equivalent capacitance CEQ is
CEQ = C1 + C2 = 16 + 4 = 20.0 µF
||
The energy stored in this combination is then
1
1 ||
W = CEQ V 2 = (20 × 10−6 )(22)2 = 4.84 × 10−3 J
2
2
For a series combination, the equivalent capacitance is
CEQ =
C1 C2
16 × 4
= 3.20 µF
=
C1 + C2
16 + 4
The voltage required to charge this combination with the same total energy is then
s
2W
V =
CEQ
r
2 × (4.84 × 10−3 )
=
3.2 × 10−6
= 55.0 V
page 8 of 8