A non conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its center:
A. decreases as r increases, for \( 0 \lt r \lt \infty \)
B. increases as r increases, for \( r \lt R \)
C. becomes zero.
D. decreases as r increases, for \( R \lt r \lt \infty \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → B and D only
For a uniformly charged solid sphere (charge \( Q \)), Gauss's law with a spherical surface of radius \( r \) gives the field inside and outside:
\( \displaystyle E = \frac{1}{4\pi\varepsilon_0}\,\frac{Q\,r}{R^{3}} \quad (r \lt R), \qquad E = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r^{2}} \quad (r \ge R) \)
Inside, the field grows linearly from zero at the centre; outside, it falls as \( 1/r^{2} \). It is largest at the surface.
(A) decreases as r increases, for \( 0 \lt r \lt \infty \) — Incorrect. Inside the sphere \( E \propto r \), so the field increases up to the surface.
(B) increases as r increases, for \( r \lt R \) — Correct. Inside, \( E \propto r \).
(C) becomes zero. — Incorrect. The field is zero only at the centre \( r = 0 \); at every other point it is non-zero, and outside it only tends to zero as \( r \rightarrow \infty \).
(D) decreases as r increases, for \( R \lt r \lt \infty \) — Correct. Outside, \( E \propto 1/r^{2} \).