Question

An electric field of 100N/C is applied to an electric dipole at an angle of 60" The value of electric dipole moment is 10-20 Cm What is the potential energy of the electric dipole?


​

Answer 1

Torque on dipole, T = P×E×Sin( ¥ )

Where

P = Dipole Moment = 2qr = q× ( 2 r )

Here q = Charge on dipole ,

2r = length of dipole = 2cm = 0.02m

E = Electric Field intensity = 1000000 N/C

¥ = angle between P and E = 60

0

Now

T = P×E×Sin ( ¥ )

= q×(2r)×E× Sin ( ¥ )

= q×2r×E× Sin ( ¥ )

q =

2r×E×Sin(¥)

T

=

0.02×1000000

8

3

×(

2

3

)

=

2×1000000×(

3

)

[8

3

)×100×2]

=

10000

8

= 8×10

−4

C

= 8×10

−4

C×

100

100

= 8×100×[

100

10

−4

]C

= 800×10

−6

C

= 800 microCoulomb