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} $