Question

Let $\rho \left(r\right)=\frac{Qr}{4\pi R^4}$ be the charge density distribution of a solid sphere of radius $R$ and total charge $Q$. Find the magnitude of electric field at a point $P$ inside the sphere at a distance $r_1$ from the centre of the sphere.

• A. $\frac{Q}{4{\pi }^2{ϵ}_0{r_1}^2}$
• B. $\frac{Q{r_1}^2}{4{\pi }^2{ϵ}_0{R}^4}$
• C. $\frac{Q{r_1}^2}{16\pi {ϵ}_0{R}^4}$
• D. $0$

Answer 1

The correct option is C $\frac{Q{r_1}^2}{16\pi {ϵ}_0{R}^4}$

Given that,

The charge density $\rho \left(r\right)=\frac{Qr}{4\pi R^4}$

According to the fig,

The electric field at the point P distant r₁ from the center,

According to gauss’s theorem

$E\times 4\pi r_1^2$= charge enclosed/ε₀

$E\times 4\pi r_1^2=\frac{1}{{\epsilon }_0}\int \rho dV$

$E\times 4\pi r_1^2=\frac{1}{{\epsilon }_0}\underset{0}{\overset{r_1}{\int }}\frac{Qr}{4\pi R^4}\times 4\pi r^{2}dr$

$E=\frac{1}{{\epsilon }_0}\times \frac{Q{r}_1^4}{4\times 4\pi R^4}\times 4\pi \times \frac{1}{4\pi r_1^2}$

$E=\frac{Q{r}_1^2}{{\epsilon }_{0}16\pi R^4}$

Therefore, the electric field at a point P will be; $E=\frac{Q{r}_1^2}{{\epsilon }_{0}16\pi R^4}$