For a uniformly charged thin spherical shell, the electric potential (V) radially away from the centre (O) of shell can be graphically represented as -

Solution

Physics sir · 3 steps

Inside the shell the field is zero, so V stays flat. Outside, the shell acts like a point charge.

1. Inside the shell. By Gauss's law there is no charge enclosed for r < R, so E = 0 everywhere inside. Zero field means the potential does not change as you move, so V stays equal to its surface value kQ/R all the way to the centre.
2. Outside the shell. For r ≥ R the shell behaves like a point charge Q at O. So V = kQ/r, which falls off as a curve (a hyperbola), not as a straight line.

   V(r) = kQ/R, r ≤ R ; kQ/r, r ≥ R

   Option	Inside (r < R)	Outside (r > R)	Matches?
   (a)	constant	falls as 1/r curve	yes
   (b)	rises from zero	straight line	no
   (c)	constant	straight line	no
   (d)	rises linearly from zero	falls as 1/r curve	no

3. So. Only graph (a) is flat inside and then drops along a 1/r curve outside, joining smoothly at r = R. The answer is (a).

The idea to keep. Wherever E = 0, the potential is constant, not zero. For any spherical charge, outside it behaves like a point charge, so V falls as 1/r.

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