Question 837: Electrostats & Capacitance

Question

The electrostatic potential inside a charged spherical ball is given by $\mathrm{V}=\mathrm{ar}^2+\mathrm{b}$ where ' r ' is the distance from its centre and ' $a$ ' and ' $b$ ' are constants. The volume charge density of the ball is [ $\varepsilon_0=$ permittivity of free space $]$

Accepted Answer

Given: $\mathrm{V}=\mathrm{ar}^2+\mathrm{b}$
$$\begin{aligned}
E & =-\frac{d V}{d r} \
& =-\frac{d}{d r}\left(a r^2+bight) \
E & =-2 a r\quad	ext{... (i)}
\end{aligned}$$
By Gauss' law,
$$\begin{aligned}
& \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~S}}=\frac{\mathrm{q}}{\varepsilon_0} \
& 4 \pi \mathrm{r}^2 \mathrm{E}=\frac{\mathrm{q}}{\varepsilon_0}\quad	ext{.... (ii)}
\end{aligned}$$
Total charge $(q)$ enclosed $=\frac{4 p}{3} \pi r^3$ ... ($ho$ = charge density of ball)
From (i) and (ii)
$$\begin{aligned}
& 4 \pi \mathrm{r}^2 	imes(-2 \mathrm{ar})=\frac{4 ho}{3 \varepsilon_0} \pi \mathrm{r}^3 \
& -2 \mathrm{a}=\frac{ho}{3 \varepsilon_0} \
& ho=-6 \mathrm{a} \varepsilon_0
\end{aligned}$$

Answer

$-24 \pi \mathrm{a} \varepsilon_0 \mathrm{r}$

Answer

$-6 \mathrm{a} \varepsilon_0 \mathrm{r}$

Answer

$-24 \pi \mathrm{a} \varepsilon_0$

Answer

$-6 \mathrm{a} \varepsilon_0$

Electrostats & Capacitance

The electrostatic potential inside a charged spherical ball is given by \mathrm{V}=\mathrm{ar}^2+\mathrm{b} where ' r ' is the distance from its centre and ' a ' and ' b ' are constants. The volume charge density of the ball is [ \varepsilon_0= permittivity of free space ]

Choose the correct answer: