Question

An insulating solid sphere of radius 'R' is charged in a non-uniform manner such that volume charge density r= $\frac{A}{r}$, where A is a positive constant and r the distance from centre. Electric field strength at any inside point at distance r₁ is –

• A. $\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$ $\frac{4\pi A}{r_{1}}$
• B. $\frac { 1 } { 4 \pi \varepsilon _ { 0 } }$ $\frac{A}{r_{1}}$
• C. $\frac{A}{\pi\varepsilon_{0}}$
• D. $\frac{A}{2\varepsilon_{0}}$

Answer 1

Answer: (D)

$\frac{A}{2\varepsilon_{0}}$

Answer 2

P is any inside point at distance r₁ from O. we take a spherical surface of radius r₁ as Gaussian –surface of radius r₁ as Gaussian-surface.

$\oint_{s}^{}{\overrightarrow{E}.\overset{\rightarrow}{ds}}$ = $\frac{q_{in}}{\varepsilon_{0}}$

By symmetry, E at all points on the surface is same and angle between $\overrightarrow{E}$and $\overset{\rightarrow}{ds}$is zero everywhere.

\ $\oint_{s}^{}{\overrightarrow{E}.\overset{\rightarrow}{ds}}$ = Es = $\frac{q_{in}}{\varepsilon_{0}}$ or E 4pr²₁ = $\frac{q_{in}}{\varepsilon_{0}}$ .....(i)

q_in : The sphere can be regarded as consisting of a large number of spherical shells.Consider a shell of inner and outer radii r and r +dr. Its volume will be dV = 4pr² dr. Charge in the shell,

dq = rdV = $\frac{A}{r}$4pr²dr

Total charge enclosed by Gaussian-surface,

q_in = $\int_{}^{}{dq}$ = $\int_{0}^{r_{1}}{rdr}$ = 4pA = $\frac{r_{1}^{2}}{2}$

q_in = 4pA $\int_{0}^{r_{1}}{rdr}$= 4pA $\frac{r_{1}^{2}}{2}$

From Eq. (1) E4p$r_{1}^{2}$ = 4pA$\frac{r_{1}^{2}}{2}$/ e₀

\ E = $\frac{A}{2\varepsilon_{0}}$