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the amount of work done in rotating an electric dipole moment3.2×10^-8cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field of intensity10^4N/C

the amount of work done in rotating an electric dipole moment3.2×10^-8cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field of intensity10^4N/C

Grade:12

2 Answers

Arun
25750 Points
4 years ago
P = 3 x 10-8cm ;
E =104N/C 
At stable equilibrium (θ1 ) = 0° 
At unstable equilibrium (θ2) =180° 
Work done in rotating dipole is given by:  W = PE (cosθ1 - cosθ2 )
  =  (3 x 10-8) (104) [cos0º - cos180º] 
= 3 x10-4 [1 - (- 1)] 
W = 6 x 10-8J
Khimraj
3007 Points
4 years ago

Work done to get the dipole in stable equilibrium is

W=pE (Cos x1–-Cos x2)

Here, x1 and x2 are the angels between dipole moment ‘p’ and electric field ‘E’ in the initial and final state of the dipole.

For unstable equilibrium, x2=180°

and for stable equilibrium, x1=0°

W=(3×10^(-8)×10^4)(Cos 0°—Cos 180°) J

=6×10^(-8) J

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