Electrostatics

JEE Main level, single correct, medium

224 of 311

Question 224 of 311

The electrostatic potential inside a charged spherical ball is given by ϕ=αρ2+b\phi=\alpha \rho^2+b where rr is the distance from the centre; a,ba, b are constants. Then the charge density inside ball is

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Answer: (C) −6aε0-6 a \varepsilon_0

Potential inside (ϕ)=ar2+b(\phi)=a r^2+b
∴Er=−δvδr=−2ar\therefore \mathrm{E}_{\mathrm{r}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{r}}=-2 \mathrm{ar}
Electric field inside uniformly charged solid volume varies with ' rr '. So charge density is constant
ϕnet =(−2ar)4πr2=−8πar3−8πar3=σ×43πr3ε0∴σ=−6aε0\begin{aligned} & \phi_{\text {net }}=(-2 \mathrm{ar}) 4 \pi \mathrm{r}^2=-8 \pi a r^3 \\ & -8 \pi \mathrm{ar}^3=\frac{\sigma \times \frac{4}{3} \pi \mathrm{r}^3}{\varepsilon_0} \\ & \therefore \sigma=-6 \mathrm{a} \varepsilon_0 \end{aligned}

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