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AIEEE

P H Y S I C S
S T U D Y M A T E R I A L

NARAYANA INSTITUTE OF CORRESPONDENCE COURSES
F N S H O U S E , 6 3 K A L U S A R A I M A R K E T
S A R V A P R I Y A V I H A R , N E W D E L H I - 1 1 0 0 1 6
PH.: (011) 32001131/32/50 • FAX : (011) 41828320
Websi te : w w w . n a r a y a n a i c c . c o m
E-mai l : i n f o @ n a r a y a n a i c c . c o m

ELECTROSTATICS

Page 33

Physics : Electrostatics

29
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INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

The potential at any arbitrary point (P3, located at r, θ coordinates) is (for r >> d)

2
0

1 pcos
V .

4 r
θ

=
πε

Potential at a point which is equidistant from +q and –q charge is zero. Thus potential at all points
lying on the equatorial plane is zero for a dipole.

POTENTIAL ENERGY OF A DIPOLE

The work done in rotating a dipole placed in a uniform electric field E, from initial angle θ1 to final angle

θ2 is θτ= ∫
θ

θ

dW
2

1

= pE (cos θ1 – cosθ2)
where U2, U1 are the potential energy of the dipole in the two orientations.
Potential energy

U2 – U1 = – pE cos θ2 – (–pE cos θ1)
= –pE (cos θ2 – cos θ1)

The zero of the potential is taken at θ = 90º . Thus, potential energy of the dipole is

→→
−= E.pU

= –pE cos θ
Umin = –pE, Umax = + pE

Work done in rotating a dipole from θ = 0 (aligned parallel to E), to θ = 180º (aligned antiparallel to

E) is W 2pE=

EXAMPLES BASED ON DIPOLE SYSTEM

Example - 33 The work required to turn an electric dipole end for end in a uniform electric field when

the initial angle between →p and

E is θ0 is -

Solution : W = pE (cosθ1 – cosθ2), in this case θ1 = θ0 and θ2 = π + θ0. Thus
W = pE {cosθ0 – cos (π + θ0)}
= 2pE cos θ0

Example - 34 Calculate the electric intensity due to a dipole of length 10 cm and having a charge
of 500 µC at a point on the axis distance 20 cm from one of the charges in air .

Solution : The electric intensity on the axial line of the dipole

222
0 )d(

pd2
4

1
E

#−∈π
=

2l = 10 cm ∴ # = 5 × 10–2 m

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INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

d = 20 + 5 = 25 cm = 25 × 10–2 m
p = 2q# = 2 × 500 × 10–6 × 5 × 10–2 ⇒ 5 × 10–3 × 10–2 = 5 × 10–5 C-m

∴ 2228
259

]525[10
10251052109

E


××××××
=



−−

⇒ 6.25 × 107 N.C.

Example - 35 Calculate the electric intensity due to an electric dipole of length 10 cm having charges
of 100 µC at a point 20 cm from each charge.

Solution : The electric intensity on the equatorial line of an electric dipole is 2/322
0 )d(

p
4

E
#+∈π

1
=

p = 2# q C-m
= 10 × 10–2 × 100 × 10–6

= 10–5 C-m
d2 + #2 = (20 × 10–2)2 = 4 × 10–2

∴ 2/32
59

)104(
10109

E




×
××

=

73
59

10
8
9

810
10109

×=
×
××

=




= 1.125 × 107 N/C

Example - 36 Find out the torque on dipole in N-m given : Electric dipole moment )k̂2ĵî5(10P 7 −+= −


coulomb metre and electric field )k̂ĵî(10E 7 ++=


Vm–1 is -

Solution :
→→→

×=τ EP = 111
215

k̂ĵî


)15(k̂)52(ĵ)21(î −+−−++= k̂4ĵ7î3 +−=

6.8|| =τ


N-m

CHARGED LIQUID DROP
If n small drops each of radius r coalesce to form a big drop of radius R, then

(i) 33 r
3
4

nR
3
4 π



∴ R = rn1/3

(ii) If each small drop has a charge q, then the charge on the big drop

nq'q =

Page 65

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INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

24 Charge are placed on the vertices of a square as shown in Fig. Let
$
E is along (2). Potential at centre

remains same
[AIEEE 2007]

(1)
$
E remains unchanged, V changes (2) Both

$
E and V change

(3)
$
E and V remain unchanged (4)

$
E changes, V remain changed

25 The potential at a poit x (measured in µm) due to some chargfes situated on the x-axis is given by

( ) ( )= −2V x 20 / x 4 volts. The electric field E at x = 4 µm is given by
[AIEEE 2007]

(1)
5
3

volt/ µm and in the -ve x direction (2)
5
3

volt/ µm and in the +ve x direction

(3)
10
9

volt/ µm and in the -ve x direction (4)
10
9

volt/ µm and in the +ve x direction

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INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

EXERCISE

LEVEL - I
1. (3) 2. (3) 3. (1) 4. (1) 5. (2)
6. (1) 7. (3) 8. (1) 9. (2) 10. (3)
11. (3) 12. (2) 13. (3) 14. (3) 15. (1)
16. (2) 17. (3) 18. (4) 19. (2) 20. (1)
21. (4) 22. (4) 23. (3) 24. (4) 25. (3)

LEVEL - II
1. (1) 2. (4) 3. (3) 4. (3) 5. (2)
6. (3) 7. (4) 8. (3) 9. (4) 10. (4)
11. (2) 12. (3) 13. (1) 14. (2) 15. (3)
16. (3) 17. (2) 18. (1) 19. (3) 20. (2)
21. (1) 22. (1) 23. (2) 24. (2) 25. (3)

LEVEL - III
1. (1) 2. (3) 3. (1) 4. (2) 5. (1)
6. (4) 7. (4) 8. (1) 9. (2) 10. (1)
11. (1) 12. (4) 13. (4) 14. (2) 15. (3)
16. (3) 17. (2) 18. (3) 19. (4) 20. (2)
21. (3) 22. (2) 23. (3) 24. (4) 25. (1)

PROBLEMS ASKED IN VARIOUS EXAMS
1. (4) 2. (1) 3. (3) 4. (1) 5. (3)
6. (1) 7. (2) 8. (4) 9. (4) 10. (2)
11. (2) 12. (2) 13. (3) 14. (2) 15. (4)
16. (1) 17. (1) 18. (3) 19. (4) 20. (4)
21. (2) 22. (2) 23. (2) 24. (4) 25. (4)

ANSWERS

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