Which of the following correctly represents the variation of electric potential (V) of a charged spherical conductor of radius (R) with radial distance (r) from the center?

Solution

Physics sir · 4 steps

Inside a conductor the field is zero, so V stays flat at its surface value; outside it falls as 1/r.

1. Inside the conductor. In electrostatic equilibrium the field inside a conductor is zero. Since E = -dV/dr, zero field means V does not change, so it stays constant from the centre up to r = R.
2. Outside the conductor. All the charge Q sits on the surface, and outside the sphere it acts like a point charge at the centre. So V falls off as 1/r.

   V(r) = kQ/R, r ≤ R ; kQ/r, r > R

3. Shape of the graph. The graph is a flat line at kQ/R from r = 0 to r = R, and then a smooth 1/r curve. The two pieces meet at r = R, because potential cannot jump.
4. So. Only graph (d) shows a constant value inside followed by a 1/r fall outside. The answer is (d).

The idea to keep. For a charged conductor, the field drops to zero inside but the potential does not. It stays equal to the surface value. Zero field means constant potential, not zero potential.

Answer from the NTA final answer key. Solution by PrepWiser.