Question

A sphere of radius $R$ carries charge density proportional to the square of the distance from the center. $\rho =A{r}^2$, where $A$ is a positive constant. At a distance of $R/2$ from the center, the magnitude of the electric field is :

• A. $A/4\pi {ϵ}_0$
• B. $A{R}^3/40{ϵ}_0$
• C. $A{R}^3/24{ϵ}_0$
• D. $A{R}^3/5{ϵ}_0$

Answer 1

The correct option is C $A{R}^3/24{ϵ}_0$

$\begin{array}{c}Accordingtoquestion..............\\ chargedensityP=\frac{chargeenclosed}{Volumeofsphere}=\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi r^3}\\ fromGaussLaw:\\ ElectricfieldEatdistancerfromcentre=\frac{Q}{A.E_0}=\frac{p.V}{A.E_0}\\ =\frac{p.\left(\frac{4}{3}\pi r^3\right)}{4\pi r^3}=\frac{pr}{3.E_0}\\ Electricfieldatdistance\frac{R}{2}fromcentre=\frac{p\left(\frac{R}{2}\right)}{3.E_0}\\ Also,\\ p=A{r}^2\\ so,pat\frac{R}{2}=A\times \frac{R^2}{4}\\ Electricfieldatdistance\frac{R}{2}fromcentre=\frac{\left[\frac{A.R^2}{4}\right]\left(\frac{R}{2}\right)}{3.E_0}=\frac{A.R^3}{24.E_0}\\ sothatthecorrectoptionisC.\end{array}$