A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in the figure. The approximate variation electric field E⃗ as a function of distance r from centre O is given by

Solution

Physics sir · 3 steps

Inside metal the field is always zero, so it vanishes in the sphere and again in the shell's thickness.

1. Charges settle. The charge q sits on the surface of the metal sphere. It induces -q on the inner surface of the shell (radius a) and +q on its outer surface (radius b).

   E · 4π r² = q_(enc)/ε₀ ⇒ E = q_(enc)/(4πε₀ r²)

   Region	Charge enclosed	Field E
   r < R (inside metal sphere)	0	0
   R < r < a (gap)	q	q/(4πε₀ r²), falling as 1/r²
   a < r < b (inside shell's metal)	q - q = 0	0
   r > b (outside)	q - q + q = q	q/(4πε₀ r²), falling as 1/r²

2. Read the graph. So E is zero up to R, jumps up and falls as 1/r² until a, is zero from a to b, then jumps again at b and falls as 1/r². The graph is only approximate, so the exact heights of the jumps do not matter.
3. So. Only graph (a) has both zero regions and both 1/r² pieces, so the answer is (a).

The idea to keep. For nested conductors, mark every region that is solid metal: E = 0 there. In every gap, use Gauss's law with the total charge enclosed, and you get a 1/r² fall-off.

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