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Question

Physics Question on Electric charges and fields

A spherically symmetric charge distribution is characterised by a charge density having the following variation : ρ(r)=ρ0(1rR)\rho\left(r\right)=\rho_{0}\left(1-\frac{r}{R}\right) for r<Rr < R ρ(r)=0\rho \left(r\right)=0 \, for rRr\ge R Where rr is the distance from the centre of the charge distribution and ρ0\rho_{0} is a constant. The electric field at an internal point (r<R)(r < R) is

A

ρ04ϵ0(r3r24R)\frac{\rho_{0}}{4\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)

B

ρ0ϵ0(r3r24R)\frac{\rho_{0}}{\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)

C

ρ03ϵ0(r3r24R)\frac{\rho_{0}}{3\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)

D

ρ012ϵ0(r3r24R)\frac{\rho_{0}}{12\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)

Answer

ρ0ϵ0(r3r24R)\frac{\rho_{0}}{\epsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4R}\right)

Explanation

Solution

ε.ds=Σaε0\oint \varepsilon . d s=\frac{\Sigma a}{\varepsilon_{0}}
ε(4πr2)=ρ0r(1rR)4πr2dr\varepsilon\left(4 \pi r^{2}\right)=\rho \cdot \int_{0}^{r}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r
4πρ0[r33r24R]0q=ε(4πr2)4 \pi \rho_{0}\left[\frac{r^{3}}{3}-\frac{r^{2}}{4 R}\right]_{0}^{q}=\varepsilon\left(4 \pi r^{2}\right)
ε=ρ0ε0(r3r24R)\varepsilon=\frac{\rho_{0}}{\varepsilon_{0}}\left(\frac{r}{3}-\frac{r^{2}}{4 R}\right)