Question

A electric dipole of length 2 cm is placed with its axis making a angle of 60° to a uniform electric field of 10^5 N/C. If it experience a torque=8√3 N-m calculate :-
1. Magnitude of charge on dipole.
2. Potential Energy of dipole.​

Answer 1

1. Magnitude of charge= 0.75 × 10^-2C.

2. Potential energy of dipole= -7.5J

Answer 2

Answer:

1.Magnitude of charge on dipole is $4\times10^{-3}C$

2.Potential energy of dipole is -8J

Explanation:

Given that,

A electric dipole of length 2cm is placed with its axis making a angle of 60° to a uniform electric field of $10^{5}N/C$ and it experiences a torque of $8\sqrt{3} N-m$.

The torque experienced by dipole is given by the relation,

$T=PEsin\theta$ where,

P=Dipole moment

E=Electric field

Substituting the given values in the above relation,we get

$8\sqrt{3} =P(10^{5})sin60= > P=16\times10^{-5}m$

The relation for dipole moment is mentioned below

$P=q\times2l$

q=Charge on dipole

l=length of dipole=2cm=0.02m

Substituting these values in the above relation,we get

$16\times10^{-5}=q\times2(0.02)= > q=4\times10^{-3}C$

Hence,the magnitude of charge on the dipole is found to be

$q=4\times10^{-3}C$

The Potential energy of dipole is given by the relation,

$U=-PEcos\theta$

Substituting known values in above relation,we get

$U=-16\times10^{-5}\times10^{5}cos60=-8J$

Therefore,the potential energy of the dipole is -8J

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