Electrostatic potential inside a charged spherical ball is given by phi = ar^2 + b where r is distan

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1. Electric Charges and Fields

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The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$ where $r$ is the distance from the centre $a,\,b$ are constants. Then the charge density inside the ball is

A

$ - \,6a{\varepsilon _0}r$

B

$ - \,24\pi a{\varepsilon _0}$

C

$ - \,6a{\varepsilon _0}$

D

$ - \,24\pi a{\varepsilon _0}r$

Solution

Electric field, $E=-\frac{d \phi}{d t}=-2 a r$

By Gauss's theorem $E\left(4 \pi r^{2}\right)=\frac{q}{\varepsilon_{0}}$

$\Rightarrow q=-8 \pi \varepsilon_{0} a r^{3}$

$\rho=\frac{d q}{d V}=\frac{d q}{d r} \times \frac{d r}{d V}$

$=\left(-24 \pi \varepsilon_{0} a r^{2}\right)\left(\frac{1}{4 \pi r^{2}}\right)=-6 \varepsilon_{0} a$

Standard 12

Physics

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