Q59P

Question

How much work is required to turn an electric dipole  180°in a uniform electric field of magnitudeE = 46.0 N/C  if the dipole moment has a magnitude of p = 3.02 ×1025C.mand the initial angle is 64°?

Step-by-Step Solution

Verified
Answer

The amount of work required to turn the electric dipole 1800 in a uniform electric field is.1.22×1023J

1Step 1: The given data

 

  1. Dipole moment,p=3.02 x 1025C.m
  2. Electric field of magnitude,E=46.0N/C
  3. Initial angle,θi=64o
  4. Final angle of the dipole,θf=180o
2Step 2: Understanding the concept of energy and dipole

The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field. The magnitude of the electric dipole moment isp=qd, where,q is the magnitude of the charge, and d  is the separation between the two charges. When placed in an electric field, the potential energy of the dipole is given by

U(θ)=p.E   =pEcos θ

Formulae:

Therefore, if the initial angle between p and E is  and the final angle isθ, then the change in potential energy would be 

    ΔU= U(θ) Uo(θ)  =pE(cos θ cos θo)                                                                          (i)

The electric dipole moment of a system, p=qd                                         (ii)  

 

3Step 3: Calculation of the amount of work

“Torque and energy of an electric dipole in an electric field,” thus we can find the amount of work can be given as the net potential difference and using equation (i) it is given as follows

W= U(θo+π) Uo(θo)= pE(cos (θo+π)  cos(θo) )= 2pEcos θo =2(3.02×1025C.m)46.0NC cos 64.0o =1.22×1023J

Hence, the value of the amount of work done for the required change in angle is.1.22×1023J