Q: An insulates sphere of radius R has a uniform volume charge density λ. The electric field at a point A, which is at a distance r from its centre, is given by : (R > r)

(a) Zero

$\displaystyle (b) \frac{R \lambda}{3 \epsilon_0} $

$\displaystyle (c) \frac{2 r \lambda}{3 \epsilon_0} $

$\displaystyle (d) \frac{r \lambda}{3 \epsilon_0} $

Ans: (d)

Sol: Charge enclosed by sphere of radius r is given by, $\displaystyle q = \frac{4}{3}\pi r^3 \lambda $ ; Since λ is the volume charge density .

According to Gauss’s Law ;

$\displaystyle \oint \vec{E}.\vec{ds} = \frac{q}{\epsilon_0} $

$\displaystyle \oint \vec{E}.\vec{ds} = \frac{4 \pi r^3 \lambda }{3 \epsilon_0} $

$\displaystyle E \times 4 \pi r^2 = \frac{4 \pi r^3 \lambda }{3 \epsilon_0} $

$\displaystyle E = \frac{r \lambda}{3 \epsilon_0} $