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ExamsJEE AdvancedPhysics

A charge +Q is fixed at the origin. A small electric dipole of dipole moment p, with its axis along the x-axis and pointing away from the charge, is released from rest from a point very far away (effectively at infinity). Find the kinetic energy of the dipole when it reaches the point (d, 0).

  1. p*Q / (4*pi*e0*d²)
  2. p*Q / (4*pi*e0*d)
  3. 2*p*Q / (4*pi*e0*d²)
  4. p*Q / (8*pi*e0*d²)

Correct answer: p*Q / (4*pi*e0*d²)

Solution

The field of +Q at distance r along the x-axis is E = Q/(4*pi*e0*r²), directed away from the charge (along +x). The dipole moment p points away from the charge (along +x) too, so the dipole is aligned with the field and its potential energy is U(r) = -p*E = -p*Q/(4*pi*e0*r²). Starting from rest at infinity where U = 0, energy conservation gives KE(d) = U(infinity) - U(d) = 0 - (-p*Q/(4*pi*e0*d²)) = p*Q/(4*pi*e0*d²).

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