Question

Let $P(r) = \frac{Q}{\pi R^{4}}r$ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r₁ from the centre of the sphere, the magnitude of electric field is –

• A. 0
• B. $\frac{Q}{4\pi \in_{0}r_{1}^{2}}$
• C. $\frac{Qr_{1}^{2}}{4\pi \in_{0}R^{4}}$
• D. $\frac{Qr_{1}^{2}}{3\pi \in_{0}R^{4}}$

Answer 1

Answer: (C)

$\frac{Qr_{1}^{2}}{4\pi \in_{0}R^{4}}$

Answer 2

P(r) =

From Gauss law

$\oint$ E.ds = = $\frac { \int \rho \mathrm { VdV } } { \varepsilon _ { 0 } }$ =

E.4pr₁² = $\frac { \mathrm { Q } } { \pi \mathrm { R } ^ { 4 } } 4 \pi \frac { \mathrm { r } _ { 1 } ^ { 4 } } { 4 }$ e₀

E = $\frac { \mathrm { Qr } _ { 1 } ^ { 2 } } { 4 \pi \varepsilon _ { 0 } \mathrm { R } ^ { 4 } }$