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?
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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
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