Module NF2FSLT SP025

(a) from : C   A   r  o A

dd

 78.85 x 10-12 2
5 x 10-3

 2.479 x 10-8 F  24.79 nF

(b) from : CQ
V

QCV

  (2.479 x 10-8 ) 10 x 103

 2.479 x 10-4 C

(c) Electric field strength, E  V
d

E  (10 x 103 )  2 x 106Vm 1
5 x 10-3

Exercise 16 LO 2.31

A parallel-plate capacitor has the space between the plates filled

with two slabs of dielectric constants 1 and 2 as shown in

Figure 2.4.

1 d/2
2 d/2

Figure 2.4

Each slab has thickness d/2, where d is the plate separation.

Show that the capacitance of the capacitor is

C  2 0 A  1 2 
1  2
d

Solution :
Since both dielectric materials are arranged in series, thus the
parallel-plate capacitor in Figure 2.7 can be considered as two
parallel-plate capacitors in series.
Therefore the total capacitance is

1 1 1
C C1 C2

1 d/2 1  1  1
C  1 0A    2 0A 
2 2
2 d/2 d d

1 d  d

C 21 0 A 2 2 0 A

Solution :  

1  d A 1  1
C
2 0 1 2

1  d   2  1 
C
2 0 A  1 2

C  20 A  1 2 
1  2
d

Additional Exercises

Exercise LO 2.1 :

Given 0 = 8.85  1012 C2 N1 m2
1. a. A parallel-plate, air-filled capacitor has circular plates

separated by 1.80 mm. The charge per unit area on each
plate has magnitude of 5.60 pC m2. Calculate the potential
difference between the plates of the capacitor.
(University physics,11th edition, Young & Freedman, Q24.4,
p.934)
b. An electric field of 2.80  105 V m1 is desired between two

parallel plates each of area 21.0 cm2 and separated by
0.250 cm of air. Determine the charge on each plate.
(Physics for scientist & engineers ,3rd edition, Giancoli, Q14,
p.628)
ANS. : 1.14 mV; 5.20  109 C

2. When the potential difference between the plates of a
capacitor is increased by 3.25 V, the magnitude of the charge
on each plate increases by 13.5 C. What is the capacitance
of this capacitor?
(Physics,3rd edition, J.S.Walker, Q86, p.694)

ANS. : 4.15 F

Additional Exercises

Exercise LO 2.1 :

3. A 10.0 F parallel-plate, air-filled capacitor with circular plates is
connected to a 12.0 V battery. Calculate
a. the charge on each plate.
b. the charge on each plate if their separation were twice
while the capacitor remained connected to the battery.
c. the charge on each plate if the capacitor were connected to the
12.0 V battery after the radius of each plate was twice without
changing their separation.

(University physics,11th edition, Young & Freedman, Q24.5, p.934)

ANS. : 120 C; 60 C; 480 C

4. A capacitor stores 100 pC of charge when it is connected across a
potential difference of 20 V. Calculate

a. the capacitance of the capacitor,

b. the amount of charge to be removed from the capacitor to
reduce its potential difference to 15 V.

ANS. : 5.0 pF; 25 pC

Additional Exercises

Exercise LO 2.1 :

Given 0 = 8.85  1012 C2 N1 m2

5. Four capacitors are connected as shown in Figure 2.5.

Figure 2.5
Calculate
a. the equivalent capacitance between points a and b,
b. the charge on each capacitor if Vab=15.0 V.

(Physics for scientists and engineers,6th edition,Serway&Jewett,
Q21, p.823)

ANS. : 5.96 F; 89.5 C on 20 F, 63.2 C on 6 F, 26.3 C on
15 F and on 3 F

Additional Exercises

6. Determine the equivalent capacitance between points a
and b for the group of capacitors connected as shown in
Figure 2.6.

Figure 2.6

Take C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F.

(Physics for scientists and engineers,6th edition,Serway&Jewett,
Q27, p.824)

ANS. : 6.04 F

Additional Exercises

7. An electronic flash unit for a camera contains a capacitor of
capacitance 850 F. When the unit is fully charged and ready for
operation, the potential difference across the plates is 330 V.
a. What is the magnitude of the charge on each plate of the fully
charged capacitor?
b. Calculate the energy stored in the “charged-up” flash unit.

(Physics,3rd edition, J.S.Walker, Q59, p.692)

ANS. : 0.28 C; 46 J

8. A parallel-plate capacitor has plates with an area of 405 cm2 and
an air-filled gap between the plates that is 2.25 mm thick. The
capacitor is charged by a battery to 575 V and then is
disconnected from the battery.
a. How much energy is stored in the capacitor?
b. The separation between the plates is now increased to
4.50 mm. How much energy is stored in the capacitor
now?
c. How much work is required to increase the separation of
the plates from 2.25 mm to 4.50 mm?

(Physics,3rd edition, J.S.Walker, Q60, p.692)

ANS. : 2.63  105 J; 5.27  105 J; 2.63  105 J







LO 3.1 Electric conduction Non-face-to face SLT

Exersice 17
There is a current of 0.5 A in a flashlight bulb for 2 min. How much
charge passes through the bulb during this time?

Solution
Given : I = 0.5 A , t = 2 min = 120 s

From: Q  I t  0.5(120) 60 C

LO 3.1 Electric conduction Non-face-to face SLT

Exercise 18

A silver wire carries a current of 3.0 A. Determine
a) the number of electrons per second pass through the wire,
b) the amount of charge flows through a cross-sectional area of the

wire in 55 s.

(Given charge of electron, e = 1.60  1019 C)

LO 3.1 Electric conduction Non-face-to face SLT

Solution 2

By applying the equation of current, I =


I  3.0 A

I  Q and Q  Ne
t
I  Ne  3.0  N 1.60 1019
t t
N  1.88 1019 electrons s1

t  55 s t

Q  It
Q  3.055 Q  165 C

LO 3.1 Electric conduction Non-face-to face SLT

Exercise 19

Explain how electrical devices can begin operating almost
immediately after you switch on, even though the
individual electrons in the wire may take hours to reach the
device.

LO 3.1 Electric conduction Non-face-to face SLT

Solution

Each electron in the wire affects its neighbors by exerting a
force on them, causing them to move.
When electrons begin to move out of a battery or source
their motion sets up a propagating influence that moves
through the wire at nearly the speed of light, causing
electrons everywhere in the wire begin to move.

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Exercise 20

A constantan wire of length 1.0 m and cross
sectional area of 0.5 mm2 has a resistivity of 4.9 ×
10−7Ω m. Find the resistance of the wire.

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Solution

Given = 1.0 ; = 0.5 2 = 0.5 × 10−6 2;
= 4.9 × 10−7 Ωm

Using: R L

A

 4.9107 (1.0)
0.5 106

R  0.98

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Exercise 21

Two wires P and Q with circular cross section are made of the
same metal and have equal length. If the resistance of wire P is
three times greater than that of wire Q, determine the ratio of their
diameters.

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Solution

Given same metal : ρP  ρQ; same length: lP  lQ

RP  3RQ Knowing that : A  πd 2
ρPlP  3 ρQlQ 4
AP AQ

4 ρPlP   4 ρQlQ   dQ2 3 dQ  3
πdP2 3 πdQ2  dP2 dP

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Exercise 22

When a potential difference of 240 V is applied across a
wire that is 200 cm long and has a 0.50 mm radius, the
current density is 7.14109 A m2. Calculate
a) the resistivity of the wire,
b) the conductivity of the wire.

LO 3.2 Ohm’s Law and resistivity Non-face-to face SLT

Solution

(a) From the definition of resistance

R  V where R  ρl and I  JA

IA

(b) The conductivity of the wire is given by

ρl  V ρ2.00  240
A JA 7.14 109

ρ  1.68 108  m

σ  1 σ  1.68 1 8  5.95 107  1 m 1
ρ 10

LO 3.2 Variation of resistance with temperature Non-face-to face SLT

Exercise 23

A platinum wire has a resistance of 0.5 Ω at 0°C. It is
placed in a water bath where its resistance rises to a final
value of 0.6 Ω. What is the temperature of the bath ?

Given : αplatinum = 3.93×10–3 –1 at 0°C
K

LO 3.2 Variation of resistance with temperature Non-face-to face SLT

Solution R  Ro [1   (T  To) ]

From:

R  Ro  Ro (T  To)

Ro (T  To)  R  Ro

T  R  Ro  To  0.6  0.5  0
0.5 (3.93 103 )
Ro 

T  50.89 C

LO 3.2 Variation of resistance with temperature Non-face-to face SLT

Exercise 24
A copper wire has a resistance of 25 m at 20 C. When the wire is carrying a
current, heat produced by the current causes the temperature of the wire to
increase by 27 C.

a) Calculate the change in the wire’s resistance.

b) If its original current was 10.0 mA and the potential difference across wire
remains constant, what is its final current?

(Given the temperature coefficient of resistivity for copper is 6.80103 C1)

LO 3.2 Variation of resistance with temperature Non-face-to face SLT

Solution

a) By using the equation for temperature variation of

resistance, thus

R  R0 1  αT 

R  R0  R0αT and R  R0  R

  R  R0αT

R  25 103 6.80 103 27

R  4.59 103 

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

b) Given = 10.0 × 10−3

By using the equation for temperature variation of
resistance, thus

R  R0 1  αT  where R  V and R V
V  V 1  αT  I I0

I I0

   1 

 I
1 1 6.80 103 27
10.0 103

I  8.45 103 A

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Exercise 25

A battery has an emf of 9.0 V and an internal resistance
of 6.0 . Determine
a) the potential difference across its terminals when it is

supplying a current of 0.50 A,
b) the maximum current which the battery could supply.

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution

(a) By applying the expression for emf, thus

  V  Ir

9.0  V  0.506.0

V  6.0 V

(b) The current is maximum when the total external resistance, R =0,

therefore

  IRr

9.0  Imax 0  6.0

Imax  1.5 A

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Exercise 26

A car battery has an emf of 12.0 V and an internal

resistance of 1.0 . The external resistor of resistance

5.0  is connected in series with the battery as shown in

figure below. V

r

R

A

Determine the reading of the ammeter and voltmeter if
both meters are ideal.

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution

By applying the equation of e.m.f., the current in the circuit is

ε  IR  r

12.0  I 5.0 1.0

I  2.0 A

Therefore the reading of the ammeter is 2.0 A.

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution

The voltmeter measures the potential difference across the
terminals of the battery equal to the potential difference across
the total external resistor, thus its reading is

V  IR

V  2.05.0

V  10 V

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Exercise 27

A battery of e.m.f 3.0 V and internal resistance 5.0  is connected to a
switch by a wire of resistance 100 . The
voltage across the battery is measured by a voltmeter. What
is the voltmeter reading when the switch is

a. off ? b. on ?

ANS : (a) 3V, (b) 2.86 V

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution

Given :   3.0 V ; r  5.0 ; R  100 

R switch



r

V

When the switch is off, no current flows (I=0) in the
circuit hence the voltmeter reads emf battery

  V  Ir

3.0  V  0

V  3.0 V

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution R

I

r

V

When the switch is on, the current flows, I is given by

  IR  r

3.0  I 100  5.0

I  2.86102 A

Therefore the voltmeter reads terminal voltage, Vt :

 Vt  IR

Vt  2.86  102 100

Vt  2.86 V

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Exercise 28

An idealized voltmeter is connected
across the terminals of a battery while the
current is varied. Figure shows a graph of
the voltmeter reading V as a function of
the current I through the battery.
Find

(a) the emf, and

(b) the internal resistance of the battery

ANS : (a) 9 V, 4.5 Ω

LO 3.4 Electromotive force (emf), internal resistance Non-face-to face SLT
and potential difference

Solution

The graph gives V = 9.0 V when I = 0 and I = 2.0 A when V = 0

(a) ξ is equal to the terminal voltage when the current is zero.
From the graph, ξ = 9.0 V.

(b) When the terminal voltage (Vt = IR) is zero, the potential drop
across the internal resistance is just equal in magnitude to the
internal emf,

ξ = Ir
r = ξ /I = (9.0 V)/(2.0 A) = 4.5 Ω

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 29 Diagram shows three
RA resistors, RA = 1Ω,
RB RB = 2Ω and RC = 3Ω. If
RC potential difference across
all resistors is 12.0 V,
calculate potential
difference for each
resistor.

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution

Concept: RA, RB and RC are connected in series, therefore,
Reff = R1 + R2 + R3

= 1+2+3 = 6 Ω
And, I = I1 = I2 = I3

From I  V  12  2A
R6

Potential difference for each resistor, V = IR
VA = IA RA = (2)(1) = 2V
VB = IB RB = (2)(2) = 4V
VC = IC RC = (2)(3) = 6V

(note: VA+VB+VC = 12V which is equal to potential difference
across all resistors)

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 30 Diagram shows three
RA resistors, RA = 1Ω,
RB RB= 2Ω and RC= 3Ω.
RC If potential difference
across all resistors is
12.0 V, calculate
current for each
resistor.

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution

 Concept: RA, RB and RC are connected in parallel, therefore,
V=VA=VB=VC=12V and I = I1 + I2 + I3

 effective resistance,

1  1  1  1  1  1  1  11  1.8333
Reff RA RB RC 1 2 3 6

Reff  6  0.5454
11

Current for each resistance,

IA  VA  12  12 A, IB  VB  12  6A, IC  VC  12  4A
RA 1 RB 2 RC 3

(Note that I=12+6+4=22A I  V  12  22.0022 A
Reff 0.5454
which is equal to

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 31 4.0  12 

2.0 
8.0 V

For the circuit above, calculate
a) the effective resistance of the circuit,
b) the current passes through the 12  resistor,
c) the potential difference across 4.0  resistor,
d) the power delivered by the battery.
Ignore the internal resistance of the battery.

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution R1 R2 R12
R3
(a) R3
V V

The resistors R1 and R2 are in series, thus R12 is

R12  R1  R2 R12  4.0  12

R12  16 
Since R12 and R3 are in parallel, therefore Reff is given by

1  1  1  1 1 9 Reff  1.7778
Reff R12 R3 16 2 16

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution

(b) Since R12 and R3 are in parallel, thus V12  V3  V  8.0 V

Therefore the current passes through R2 is given by

I2  V12 I2  8.0
R12 16
I2  0.50 A

(c) Since R1 and R2 are in series, thus I1  I2  0.50 A

Hence the pVo1tenIti1aRl1differenVc1eac0r.o5s0s4R.01 is

V1  2.0 V P V2 P  8.02
(d) The power delivered by the battery is
Reff 1.78

P  36.0 W

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Exercise 32

20 
A

10  10  5.0 
B 5.0 

For the circuit above, calculate the effective resistance
between the points A and B.

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution

R1  5.0  R2  5.0  R3  10  R4  20  R5 10 

A R4 A R4

R5 R3 R1 R5 R3 R12

R2 B
B

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution

R1 and R2 are connected in series, thus R12 is

R12  R1  R2 R12  5.0  5.0  10 

A R4 Since R12 and R3 are connected in
R123 parallel , thus R123 is given by
R5
1  1 1 1 11
B R123 R12 R3 R123 10 10

A R123  5.0 

R5 R1234 R123 and R4 are connected in series , thus
R1234 is given by
B
R1234  R123  R4 R1234  5.0  20

R1234  25 

LO 3.5 Resistors in series and parallel Non-face-to face SLT

Solution A

Reff

B

Since R1234 and R5 are connected in parallel , therefore the effective
resistance Reff is given by

1  1 1
Reff R1234 R5
1 11
Reff 25 10
Reff  7.14 