Module NF2FSLT SP025

CHAPTER 1:
ELECTROSTATIC

Electrostatics is a NF2F
branch of physics
that studies electric SP025
charges at rest.
PHYSICS UNIT

KOLEJ MATRIKULASI JOHOR

1

Exercise 1 LO1.1

Figure shows three point charges that lie in the x, y plane in a vacuum. Find the electrostatic force on q1 .
q2

q1 q3 Step 2 : Find magnitude of electric force
Solution Step 1 : Draw the electric force vectors
The magnitude of the forces are
F12 sin 73
q1 F12  k Q1Q2  (9109 ) (4106 )(6106 )
r2 (0.15) 2
F12 cos 73
12

F12  9.6 N

F 13  k Q1Q3  (9109 ) (4106 )(5106 )
r2 (0.10) 2

13

F13  18 N 2

Step 3 : Adds as vector (consider the direction)

Force x - comp y - comp
F12 +9.6cos73 +9.6sin73
= +2.8 N = +9.2 N
F13
F + 18 N 0N

Fx = + 21 N Fy = +9.2 N

The electrostatic force acting on q1 :

F  Fx 2  Fy 2  212  9.2 2  23N

tan   Fy  9.2    24 above  x
Fx 21

3

Exercise 2 LO1.1

A 2 C charge lies on the straight line between a 3 C charge and a 1 C charge. The separation between the 3
C and 1 C is 4 cm.
a) Draw the position of the three charges and show the forces acting on the 2 C charge.
b) Calculate the distance of 3 C from 2 C where net force on 2 C is zero.

Solution 4 102 m
(a)
x 4 102  x
Q1  3106 C  q  2 106 C F2
F1 Q2  1106 C

(b) Nett force acting on q is zero,

F1  F2

3106 1106
x2 4102  x 2
 kQ1q  kQ2 q  
r12 r2 2

x  2.53102 m

4

Exercise 3 LO1.1

A small point charge of mass 80 g and charge of +0.600 C is hung by a thin wire of negligible mass. A charge
of 0.900 C is held 15.0 cm away from the first charge and directly to the right of it, so the wire makes an

angle  to the vertical as shown in Figure.

a. Sketch a free body diagram of the charge +0.600 C.
b. Calculate

i. the angle  ,

ii. the tension in the wire.
(Given electrostatic constant, k = 9 109 N m2 C2 and gravitational acceleration, g = 9.81 m s2)

Solution Q1  0.6 106 C; Q2  0.9 106 C;

m1  80 103 kg; r  15.0 102 m;

a. The free body diagram of the charge is

T T cos



  15.0 cm –
T sin  + Q1 F
+ 0.9μC

m1 g 0.6 μC 5

b. i. Since the charge, Q1 is in equilibrium, thus ii. By substituting  = 15.4 into eq. (2), hence

Therefore  F  0  T cos15.4  80 103 9.81

 Fx  0  T 
80 103 9.81
T sin  F
cos15.4
kQ1Q2
T sin   r2 (1)
(2)
and  Fy  0 T  0.814 N

T cos  m1g

(1) (2): T sin   kQ1Q2
T cos m1gr 2

       tan 
9.00 109 0.600 106 0.900 106

80 103 9.81 15.0 102 2

  15.4

6

Exercise 4 LO1.2
Two point charges, Q1 = +7 μC and Q2 = -5 μC are separated by a distance of 0.3 m between each other as in
figure below. Determine the resultant E produced by these two charges at point P.

P

Solution (1) Draw the electric field, E diagram produced by Q1 and Q2 at point P.
0.4 m

E1 E1 is produced by Q1 and E2 is produced by Q2. + −
Q2
P Q1 0.3 m
7
0.4 m  Use this concept :
E is outward for + charge
+ E2 E is inward for − charge
Q1 0.5 m
Vector E is draw along the line that
− joining point P and the charge.
Q2
0.3 m

(2) Apply = to find the magnitude of the each electric field create the surrounding point charge(s)
2

E1  kQ1
r12

  9109 7106

E1  0.42  3.93105 N C1
Left out the (−) sign of

E2  kQ2 the charge for Q2.
r22 Substitute the magnitude

of the charge only

E2  9109 5106   1.80105 N C1

0.52

8

(3) Adds vectorially (consider the direction) to get the resultant.

E1 Given that ; Since E is a vector quantity,

E2x tan θ = 0.4 / 0.3 so we have to resolve E1 and
Pθ θ = 53.13° E2 into x and y component
E2y E2 Resolve E2 into comp x & y and find the summation of

each of the component.

Vector x-comp.(N C-1) y-comp.(N C-1)

E1 0 3.93105

 1.80105 sin 53.13
E2 1.80105 cos 53.13
 1.08105  1.44 105

 Ex  1.08 105 N C 1

 Ey  2.49 105 N C 1

9

Therefore, the magnitude of the resultant E is ;

Ex2  Ey2
 E 

E  (1.08 105 )2  (2.49 105 )2

E  2.714 105 N C 1

Direction of the resultant E is given by ;

tan   Ey   2.49 105 
 1.08 105 
Ex  

  66.64 above the positive x-axis

10

Exercise 5 LO1.2

Two point charges, Q1= 3.0 C and Q2= 5.0 C, are placed 12 cm and 30 cm from the point P respectively

as shown in Figure below. Q- 1 P Q2
-

12 cm 30 cm

Determine

a) the magnitude and direction of the electric field intensity at P,

b) the nett electric force exerted on qO= +1 C if it is placed at P,
c) the distance of a point from Q1 where the electric field intensity is zero.

Solution  P (1) Draw the -
a. E1P E vectors
Q2
Q1 r1 
E2P

r2

By applying the equation of electric field intensity, thus Left out the (−) sign of

Find   kQ1 the charge. Substitute the
magnitude E E1P r12 magnitude of the charge
only
    E1P 
9.00 109 3.0 106  1.88 106 N C 1 11
12 102 2
Direction : to the left (towards Q1)

Left out the (−) sign of

  kQ2 the charge. Substitute the
E2P r2 2 magnitude of the charge
only

E2P    9.00 109 5.0 106  5.00 105 N C 1 Direction : to the right (towards Q2)
 30 102 2

Therefore the electric field intensityat P is given by 
EP  E1P  E2P
(3) Add E as vectors

EP   E1P  E2P

EP  1.88 106  5.00 105

EP  1.38 106 N C1

Direction : to the left (towards Q1)

12

b) From the definition of electric field intensity,

EP  FP
q0

The nett electric force exerted on qo is given by

FP  q0 EP Q1  A  Q2
E1A E2A
c. - -

  FP  1.0 106 1.38 106

x r12  x

FP  1.38 N r12

Direction : to the left (towards Q1) The electric field at A is zero, hence


E1A  E2A

To get electric field at A zero

means E1A and E2A must have kQ1  kQ2
same magnitude but point in r1A 2 r2 A 2

opposite direction

 3.0 106 
5.0 106

x 2 42 102  x 2

x  0.183 m 13

Exercise 6 LO1.3

Two point charges, Q1= 40 C and Q2= 30 C are separated by a distance of 15 cm as shown in Figure below.

Calculate B

a. the electric potential at point A and describe the meaning of the answer, 13 cm
b. the electric potential at point B.

Q1 A 10 cm Q2

- -

5 cm

Solution Q1   40  10 6 C ; Q 2   30  10 6 C Q1 Q2
r1A  5  1 0 2 m ; r2 A  1 0  1 0 2 m
-A -
a. The electric potential at point A is r2A
r1A
VA  V1A  V2A

VA   kQ1  kQ2  Substitute sign VA  9.9  106 V
 r1A  r2 A  of charge in
   equation 9.89  106 joule of work is done by the
electric field in bringing 1 C of positive
 9  109  40 106  30 106  charge from infinity to the point A. 14
 5 102 10 102 
 

Exercise 7 LO1.3

Two points, S and T are located around a point charge of +5.4 nC as shown in Figure below. Calculate

a. the electric potential difference between points S and T, S 8.0 cm T
b. the work done in bringing a charge of 1.5 nC from point T to point S.

Solution 6.0 cm

a. The electric potential difference between S and T is given by

VST  VS  VT

VST  kQ  kQ  5.4 nC
rS rT

VST  kQ 1 1  b. WTS  qVST
rS rT  (1.5109 )324

   9.00109  1 1  WTS  4.86107 J
5.4 109  6.0 102  10 102 

VST  324 V

• Work is done by the system (electric force), W negative
• Potential energy, U decreases

16

Exercise 8 LO1.3

A test charge qO =+2.3 C is placed 20 cm from a point charge Q. A work done of 25 joule is required in bringing
the test charge qO to a distance 15 cm from the charge Q. Determine
a. the potential difference between point 15 cm and 20 cm from the point charge, Q,
b. the value of charge Q,
c. the magnitude of the electric field strength at point 10 cm from the charge Q.

Solution A  q0
a. The electric potential difference between B and A is given by

QB

rB

rA

WAB  qVBA

25  2.3106 VBA

VBA  1.09 107 V 17

b. By applying the equation of electric potential difference, thus

VBA  VB  VA

VBA  kQ 1  1 
rB rA

 1.09 107   1 1 
9 109 Q  15 102  20 102 

Q  7.27 104 C

c. The magnitude of the electric field strength is E  kQ
r2

  E 
9.00 109 7.27 104

0.102

E  6.54 108 N C1

18

Exercise 9 LO1.3

(1) Two point charges, Q1= +2.0 C and Q2= 6.0 C, are placed 4.0 m and 5.0 m from a point P

respectively as shown in Figure below. Q2 -
a) Calculate the electric potential at P due to the charges.

b) If a charge Q3= +3.0 C moves from infinity to P, determine 5.0 m

the change in electric potential energy for this charge.

c) When the charge Q3 at point P, calculate the electric potential P

energy for the system of charges. Q1 + 4.0 m

Solution

a. The electric potential at point P is given by

VP  V1P  V2P

VP  k  Q1  Q2 
r1P r2 P

   2.0 106 6.0 106 
9.00 109   5.0 
 4.0 

VP  6300 V 19

b. U  qV

U  Q3 VP  V  *V  0

U  3.0 106  6300  0

U  1.89 102 J

-c. Q2 U  U12  U13  U 23

r12  3.0 m r23  5.0 m U  kQ1Q2  kQ1Q3  kQ2Q3
r13  4.0 m r12 r13 r23
Q1 +
P

+ Q3

    2.0 106 6.0 106    2.0 106 3.0 106 

 9.00 109   4.0 
 3.0
    
Substitute sign of 6.0 106 3.0 106 
5.0
charge (+ or –)

in equation U  5.49 102 J 20

Exercise 10 LO1.4

Figure above shows an electron entering a charged parallel plates with a speed of 5.45  106 m s1. The electric
field produces by the parallel plates has deflected the electron downward by a distance of 0.618 cm at the point
where the electron exits. Determine

a. the magnitude of the electric field,

b. the speed of the electron when it exits the parallel plates. 
(Given e=1.60  1019 C and me=9.11  1031 kg) u

Solution

a. The components of the initial velocity for electron are t  4.13109 s

ux  u  5.45106 m s1and uy  0

The time taken by the electron travels from one end to another

end of the plates is given by sx  uxt 2.25102  5.45106 t

Therefore the magnitude of the electric field is sy  u yt  1 ayt 2 and ay  qE
2 m

1  1.601019 E 
2 9.111031
  0.618102 2 E  4126 N C1
  4.13109
21

b. The components of the final velocity for electron are

vx  ux  5.45106 m s1
 qE
vy  uy  ayt and ay m

vy  0   qE t 2
 m 

 1.60 1019 4126 

9.111031
 vy
 4.13109

Therefore the final speed of the electron in the uniform electric

field is vy  2.99106 m s1

   v  vx2  vy2 v  5.45106 2   2.99106 2

v  6.22106 m s1

22

CHAPTER 2

CAPACITOR AND DIELECTRIC

NON FACE TO FACE SLT

SUBTOPICS :
2.1 Capacitance and capacitors in series and

parallel
2.2 Charging & Discharging of capacitors
2.3 Capacitors with dielectrics

Overview:

Capacitors and dielectrics

Capacitance Capacitors Charging Dielectric
and discharging

of capacitor

In series In parallel

Exercise 11 LO 2.11

A circular parallel-plate capacitor with radius of 1.2 cm is connected to a 6.0
V battery. After the capacitor is fully charged, the battery is disconnected
without loss of any of the charge on the plates. If the separation between
plates is 2.5 mm and the medium between plates is air.

a. Calculate the amount of charge on each plate.

If their separation is increase to 8.0 mm after the battery is disconnected,
determine

b. the amount of charge on each plate.
c. the capacitance of the capacitor.
d. the potential difference between the plates.
(Given permittivity of free space, 0 = 8.85  1012 C2 N1 m2)

Solution : r  1.2 102 m;V  6.0 V; d  2.5103 m

a. The area of each plate is

 A  πr 2 A   1.2102  4.52104 m2

The amount of charge on each plate is given by Q  CV and C  ε0 A
d
Q  ε0 AV
d
  Q 
8.851012 4.52 104 6.0
2.5103

Q  9.60 1012 C

Given d1= 8.0103 m Q  9.60 1012 C
b. The amount of each plate remains unchanged i.e.

because no charge losses to the surrounding.

Solution :
c. The new capacitance of the capacitor is

C1  ε0 A
d1

  8.851012 4.52104

 8103

C1  5.0 1013 F

d. Apply: Q  C1V1

 9.601012  5.01013 V1

V1  19.2 V

Exercise 12 (a) LO 2.11

What is the total capacitance in a, b and c ?

+- +-

+ +- - + -+ +- -
+- +-+ -
+-
+-
(a)
(b) (c)

(a) In parallel , the total capacitance is given by
(b) In series, the total capacitance is given by

(c) The equivalent capacitor for the 2 capacitors
connected in parallel :

The total capacitance for the parallel & series
connections is CE where

Exercise 12 (b) LO 2.11 D

A C1

B C2
C3

Figure 2.1

In Figure 2.1, C1= 100 F, C2 = 200 F and C3 = 300 F. The
applied potential difference between points A and B is VAB = 8.0 V.

Calculate
a. the charge on each capacitor.
b. the potential difference across each capacitor.
c. the potential difference between points A and D.

Solution : C1  100 F; C2  200 F;C3  300 F;VAB  8.0 V

A C1 C12

A

C2 D D
C3
C3

BB

a. Capacitors C1 and C2 are connected in parallel then C12 is
C12  C1  C2 C12  100 μF  200 μF
μF
Thus the effective capacitaCnc1e2  300 circuit is given by
the
Ceff in
1  1 1 1 11
Ceff C12 C3 Ceff 300 300

Ceff  150 μF

Solution : C1  100 F; C2  200 F;C3  300 F;VAB  8.0 V
a. The total charge Q stored in the effective capacitance Ceff is

Q  Ceff VAB Q  150 μ8.0

Q  1200 μC

Since the capacitors C12 and C3 are connected in series then the

charge stored in each capacitor is the same as the total charge.

Q3  Q12  Q  1200 μC

The potential difference across the capacitor C3 is

V3  Q3 V3  1200 
C3 300 

V3  4.0 V

thus the potential difference across the capacitor C12 is given by

V12  VAB  V3
V12  8.0  4.0

V12  4.0 V

Solution : C1  100 F; CC12 an2d0C0 2Fa;rCe 3con3n0e0ctedF;iVnApBara8ll.e0l V

a. Since the capacitors then

the potential difference across each capacitor is the same as V12.

V1  V2  V12  4.0 V
Q1  C1V1  100  4.0
 Therefore

Q1  400 C
and Q2  Q12  Q1

Q2  1200  400
Q2  800 C

b. The potential difference across each capacitor is given by

V1  V2  V3  4.0 V

c. The potential difference between points A and D is given by

VAD  V12  4.0 V

Exercise 13 (a) LO 2.11

a) Find the potential difference across the
capacitors X, Y and Z .

b) Find the charges reside on the capacitors X,
Y and Z .

The equivalent capacitance for Y & Z is CYZ
where

The equivalent capacitance for X, Y & Z :

The total charge for the equivalent
capacitance is

The charge for the capacitor X = 2.4x10-5 C
Potential difference across X ,

Y & Z is wired in parallel, so
Thus :

Exercise 13 (b) LO 2.11

20 V

Figure 2.2 C1
C2 C3

Figure 2.2 shows a combination of three capacitors where C1= 100 F,
the
cCo2mb=in2a2tionF. DaentdermC3ine= 47 F. A 20 V supply is connected to

a. the effective capacitance in the circuit,

b. the charge stored in the capacitor C1,
c. the potential difference across the capacitor C2,
d. the energy stored in the capacitor C3,
e. the area of the each plate in capacitor C1 if the distance

between two plates is 0.02 m and the region between plates is

vacuum.

(Permittivity of free space, o = 8.85  1012 C2 N1 m2)

Solution : C1  100 μF; C2  22 μF;C3  47 μF;V  20 V

VV

C1 C1

C2 C3 C23

a. Capacitors C2 and C3 are connected in series then C23 is
1 11
C23 C2 C3 1 11
C23 22 47
μF
Therefore the effective capacitCan23ce, C1e5ff.0is by
given

Ceff  C1  C23 Ceff  100 15.0
Ceff  115 μF

bS.oSluinticoent:hCe 1capa1c0i0toμrsFC; C1 a2 nd2C22μ3 Far;eCc3onn4e7cteμdFi;nVpar2al0leVl,
thus
Hence the charVg1e stoVr2e3d  V  20 V C1 is

in the capacitor

Q1  C1V1 Q1  100 μ20

Q1  2000 μC

c. The total charge stored in the circuit is given by
Q  115 μ20
Q  Ceff V
Q  2300 μC

Thus the charge stored in2t3h0e0caμpacit2o0r 0C023μis Q23
Q23  300 μC
Q  Q1  Q23

The capacitors C2 anQd 3C3 aQre23conn3e0ct0edμiCn series, thus

Q2

Solution : C1  100 μF; C2  22 μF;C3  47 μF;V  20 V
c. Therefore the potential difference across is
300 μ the capacitor C2
22 μ
V2  Q2 V2 
C2

V2  13.6 V

d. The energy stored in the capacitor C3 is given by

1 Q32  U3 2
2 C3 1 300 106
U3  2 47 106

U3  9.58104 J

e. By applying the capacitance’s equation for parallel-plate

capacitor, thus  100106 

C1  0A 8.851012 A
0.02 106
d

A  0.226 m2

Exercise 14 (a) LO 2.21

An uncharged capacitor and a resistor are
connected in series to a battery. If V = 12.0 V,
C = 5.0 μF & R = 8.0x105 Ω. Find
(a) the time constant
(b) the maximum charge on the capacitor
(c) the maximum current in the circuit
(d) the charge & current as functions of time
(e) the charge on the capacitor after one time
constant has elapsed.

Solution
(a)
(b) The maximum charge on the capacitor :

(c) The maximum current in the circuit :

(d) Current as a function of time
Charge as a function of time

(e) t = г = 4 s , Q = ?

Exercise 14 (b) LO 2.11

Consider the circuit shown in Figure 2.3, where C1= 50 F, C2 = 25 F
and V = 25.0 V.

S1 S2

C1 C2

V

Figure 2.3

Capacitor C1 is first charged by closing a switch S1. Switch S1 is then

opened, and then the charged capacitor is connected to the uncharged

capacitor C2 by closing a switch S2. Calculate the initial charge acquired
by C1 and the final charge on each capacitor.

Solution : C1  50 μF; C2  25 μF;V  25.0 V
Switch S1 is closed:

When the capacitor C1 is fully charged, the charge has been

placed on its plate is given by
Q1  50 25.0
Q1  C1V
Switch S2 is closed and S1 isQop1 ene1d2:50 μC

The capacitors C1 and C2 (uncharged) are connected in parallel

and the equivalent capacitance is

Ceq  C1  C2 Ceq  50  25
Ceq  75 μF
By using the principle of conservation of charge, the total charge

Q on the circuit is given by

Q  Q1  Q2 Q  1250  0
Q  1250 μC

STohleuptiootnen: tCia1ldi5f0feμreFn;cCe2acro2s5sμeFa;cVh  25.0 V is the same

capacitor

(parallel connection) and given by

V' Q V ' 1250 
Ceq 75 

V ' 16.7 V
Therefore the final charge accumulates on the

capacitor C1 : Q1'  C1V '

Q1' 50 16.7

Q1' 835 C

capacitor C2 : Q2 ' Q  Q1' OR Q2 ' C2V '

Q2 ' 1250  835
Q2 ' 415 μC

Exercise 15 (a) LO 2.21

Exercise 15 (a)

A parallel plate capacitor consists of 2 plates each with area 200 cm2
separated by a 0.4 cm air gap.
(a) compute its capacitance
(b) if the capacitor is connected across a 500 V source, what are the

charge on either plate ?

Solution Air gap – εr = 1.0 thus ε = ε0
A = 200 cm2 = 200x10-4 m2

d = 0.4 cm = 0.4x10-2 m

(a) from : C   0 A

d

 8.85x1012 (200 x1 0 4 )
0.4 x102

 44 x10 12 F  44 pF

(b) from : C  Q
V

QCV
 (4.4 x10 11) 500
 22 x10 9 C

Exercise 15 (b) LO 2.21

A parallel plate capacitor is constructed using mica as the dielectric material. The

distance between the parallel plates of the capacitor is 5.0 mm and the area of

each plate is 2.0 m2 . A potential difference of 10 kV is applied across the plates,

find (εr for mica = 7.0)

(i) The capacitance

(ii) The charge on each plate

(iii) The electric field strength between the plates

Solution

εr = 7.0 , ε = εrε0
A = 2.0 m2
d = 5 x 10-3 m
V = 10 x 103 V