Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius R, with distance r from the centre O is represented by:

Solution

Physics sir · 3 steps

Inside, the enclosed charge grows as r³, so E grows as r. Outside, the sphere acts like a point charge.

1. Inside the sphere. Take a Gaussian sphere of radius r < R. It encloses only the fraction r³/R³ of the total charge Q, so E · 4π r² = (Q r³)/(ε₀ R³). This gives E = (Q r)/(4πε₀ R³), a straight line through the origin.
2. Outside the sphere. For r > R the Gaussian sphere encloses all of Q, so E = Q/(4πε₀ r²). This is a curve that falls as 1/r², the same as for a point charge at O.

   E(r) = (Q r)/(4πε₀ R³), r ≤ R ; Q/(4πε₀ r²), r ≥ R

3. So. Both formulas give Q/(4πε₀ R²) at r = R, so the graph peaks at the surface: a straight-line rise from zero, then a 1/r² fall. That is graph (d).

The idea to keep. For a uniformly charged solid sphere, E ∝ r inside and E ∝ 1/r² outside, with the maximum at the surface. Find the enclosed charge first, and the shape of the graph follows.

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