Question 273: Electrostats & Capacitance

Question

The electrostatic potential in a charged spherical region of radius $r$ varies as $V=a r^3+b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha 	imes \pi a \in_0$. The value of $\alpha$ is $\_\_\_\_$ .
(permittivity of vacuum is $\epsilon_0$ )

Accepted Answer

The electric field $\vec{E}$ is related to the electric potential $V$ by $E=-\frac{d V}{d r}$.
Since the potential V depends only on the radial distance r , the electric field is radial.
Given the potential function :
$$ \mathrm{V}=\mathrm{ar}^3+\mathrm{b} $$
Then the electric field is,
$$ \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}}=-\left[\frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{ar}^3+\mathrm{b}ight)ight]=-3 \mathrm{ar}^2 $$
We need to find the total charge enclosed in a sphere of unit radius ( $\mathrm{R}=1$ ).
The electric flux passing through the surface of this sphere is given by $\Phi_E=\oint \vec{E} \cdot d \vec{A}$
At $R=1$, the electric field magnitude is :
$$ E(1)=-3 a(1)^2=-3 a $$
The surface area of the sphere is $4 \pi R^2=4 \pi(1)^2=4 \pi$.
Thus, the flux is :
$$ \Phi_{\mathrm{E}}=\mathrm{E} 	imes 	ext { Area }=(-3 \mathrm{a}) 	imes(4 \pi)=-12 \pi \mathrm{a} $$
According to Gauss's Law,
$$ \Phi_{\mathrm{E}}=\frac{\mathrm{Q}_{	ext {enclosed }}}{\epsilon_0} \Rightarrow \mathrm{Q}_{	ext {enclosed }}=\Phi_{\mathrm{E}} 	imes \epsilon_0 $$
$$\Rightarrow $$ $Q_{	ext {enclosed }}=(-12 \pi a) 	imes \epsilon_0$
$$\Rightarrow $$ $\mathrm{Q}_{	ext {enclosed }}=-12 \pi \mathrm{a} \epsilon_0=\alpha 	imes \pi \mathrm{a} \epsilon_0 \Rightarrow \alpha=-12$

Answer

-12

Answer

-8

Answer

-9

Answer

-6

Electrostats & Capacitance

The electrostatic potential in a charged spherical region of radius r varies as V=a r^3+b, where a and b are constants. The total charge in the sphere of unit radius is \alpha \times \pi a \in_0. The value of \alpha is \_\_\_\_ . (permittivity of vacuum is \epsilon_0 )

Choose the correct answer: