Question 528: Electrostatics & Capacitance

Question

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:

Accepted Answer

Electric field
$$E = {{d\phi } \over {dr}} =  - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i ight)$$
By Gauss's theorem
$$E = {1 \over {4	heta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$$
From $$\left( i ight)$$ and $$\left( ii ight),$$
$$q =  - 8\pi {\varepsilon _0}a{r^3}$$
$$ \Rightarrow dq =  - 24\pi {\varepsilon _0}ar{}^2dr$$
Charge density, $$ho  = {{dq} \over {4\pi {r^2}dr}} =  - 6{\varepsilon _0}a$$

Answer

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

Answer

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

Answer

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

Answer

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

Electrostatics & Capacitance

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:

Choose the correct answer: