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 $r_{1}$ from the centre of the sphere, the magnitude of the electric field is

• A. Zero
• B. $\frac{Q}{4 \pi \, \epsilon _{0 \, } r_{1}^{2}}$
• C. $\frac{Q \, r_{1}^{2}}{4 \pi \, \epsilon _{0 \, } R^{4}}$
• D. $\frac{Q \, r_{1}^{2}}{3 \pi \, \epsilon _{0} \, R^{4}}$

Answer 1

Answer: (C)

$P\left(\right.r\left.\right)=\frac{Q}{\pi R^{4}}r$

$\oint E \cdot ds =\frac{ q _{ en }}{\varepsilon_{0}}=\frac{\int \rho VdV }{\varepsilon_{0}}$
$=\frac{ Qr 4 \pi r ^{2} dr }{\pi R ^{4}} / \varepsilon_{0}$
E. $4 \pi r_{1}^{2}=\frac{\frac{Q}{\pi R^{4}} 4 \pi \frac{r_{1}^{4}}{4}}{\varepsilon_{0}}$
$E =\frac{ Qr _{1}^{2}}{4 \pi \varepsilon_{0} R ^{4}}$