Example – 1 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 q0 is
Solution : W = pE (cosq1 – cosq2), in this case q1 = q0 and q2 = p + q0. Thus
W = pE {cosq0 – cos (p + q0)}
= 2pE cos q0
Example – 2 Calculate the electric intensity due to a dipole of length 10 cm and having a charge
of 500 mC 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
E = (1/4pÎ0)(2pd/(d2-e2)2)
2l = 10 cm \ e = 5 x 10-2 m
d = 20 + 5 = 25 cm = 25 × 10-2m
p = 2qe = 2 × 500 × 10-6 × 5 × 10-2 Þ 5 × 10-3 × 10-2 = 5 × 10-5C-m
Example – 3 Calculate the electric intensity due to an electric dipole of length 10 cm having charges
of 100 mC at a point 20 cm from each charge.
Solution : The electric intensity on the equatorial line of an electric dipole is
p = 2e q C-m
= 10 × 10-22 × 100 × 10-6
= 10-5 C-m
D2 + e<2 = (20 × 10-2)2 = 4 × 10-2
Example – 4 Find out the torque on dipole in N-m given : Electric dipole moment ®p = 107 (5i^ + j^ - 2k^)coulomb metre and electric field ®E = 107(i^+j^+k^) Vm-1
Solution: 
= i^(1+2)+ j^(-2-5) + k^(5-1) = 3i^+7j^+4k^
|®t| =8.6 N-m
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