JEE MAIN - Physics (2011 - No. 12)
The electrostatic potential inside a charged spherical ball is given by $$\phi = a{r^2} + b$$ where $$r$$ is the distance from the center and $$a,b$$ are constants. Then the charge density inside the ball is:
$$ - 6a{\varepsilon _0}r$$
$$ - 24\pi a{\varepsilon _0}$$
$$ - 6a{\varepsilon _0}$$
$$ - 24\pi {\varepsilon _0}r$$
Explanation
Electric field
$$E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
By Gauss's theorem
$$E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
From $$\left( i \right)$$ and $$\left( ii \right),$$
$$q = - 8\pi {\varepsilon _0}a{r^3}$$
$$ \Rightarrow dq = - 24\pi {\varepsilon _0}ar{}^2dr$$
Charge density, $$\rho = {{dq} \over {4\pi {r^2}dr}} = - 6{\varepsilon _0}a$$
$$E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
By Gauss's theorem
$$E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
From $$\left( i \right)$$ and $$\left( ii \right),$$
$$q = - 8\pi {\varepsilon _0}a{r^3}$$
$$ \Rightarrow dq = - 24\pi {\varepsilon _0}ar{}^2dr$$
Charge density, $$\rho = {{dq} \over {4\pi {r^2}dr}} = - 6{\varepsilon _0}a$$
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