Question
The electric potential inside a charged spherical ball is given by $\phi=a r^2+b$. Where $r$ is the distance from the centre, $a$ and $b$ are constants. Calculate the charge density inside the ball.
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Answer
$\phi=a r^2+b$
$E =\frac{d \phi}{d t}=-2 a r$
$\oint \overrightarrow{ E } \cdot d \overrightarrow{ S }=\frac{q}{\epsilon_0}$
$-2 a r \cdot 4 \pi r^2=\frac{q}{\epsilon_0}$
$a=-8 \pi \epsilon_0 a r^3$
charge density $\rho=\frac{q}{\frac{4}{3} \pi r^3}=-\frac{8 \pi \epsilon_0 a r^3}{\frac{4}{3} \pi r^3}$
$\rho=-6 a \epsilon_0$ Ans.
$E =\frac{d \phi}{d t}=-2 a r$
$\oint \overrightarrow{ E } \cdot d \overrightarrow{ S }=\frac{q}{\epsilon_0}$
$-2 a r \cdot 4 \pi r^2=\frac{q}{\epsilon_0}$
$a=-8 \pi \epsilon_0 a r^3$
charge density $\rho=\frac{q}{\frac{4}{3} \pi r^3}=-\frac{8 \pi \epsilon_0 a r^3}{\frac{4}{3} \pi r^3}$
$\rho=-6 a \epsilon_0$ Ans.
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