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A spherical conductor of radius $12 \mathrm{~cm}$ has a charge of $1.6 \times 10^{-7} \mathrm{C}$ distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point $18 \mathrm{~cm}$ from the centre of the sphere?
(a) inside the sphere
(b) just outside the sphere
(c) at a point $18 \mathrm{~cm}$ from the centre of the sphere?
Solution:
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Verified Answer
Given, $\mathrm{R}=12 \mathrm{~cm}=0.12 \mathrm{~m}, \mathrm{q}=1.6 \times 10^{-7} \mathrm{C}$
(a) $\mathrm{r} < 12 \mathrm{~cm}, \mathrm{E}=$ ?
(b) $\mathrm{r}=\mathrm{R}=12 \mathrm{~cm}, \mathrm{E}=$ ?
(c) $r=18 \mathrm{~cm}, \mathrm{E}=$ ?
(a) For a point inside the spherical conductor, electric field, $\mathrm{E}=$ zero
(b) For point just on the surface of the conductor, E
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{R}^2} \\
&=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2}=\frac{9 \times 1.6 \times 10^2}{144 \times 10^{-4}} \\
&=10^5 \mathrm{NC}^{-1}
\end{aligned}
$$
$=10^5 \mathrm{NC}^{-1}$. For point outside the conductor,
(c) For point outside the conductor,
$$
\begin{aligned}
\mathrm{E} &=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} \\
&=\frac{9 \times 1.6 \times 10^2}{324 \times 10^{-4}}=4.4 \times 10^4 \mathrm{NC}^{-1} .
\end{aligned}
$$
(a) $\mathrm{r} < 12 \mathrm{~cm}, \mathrm{E}=$ ?
(b) $\mathrm{r}=\mathrm{R}=12 \mathrm{~cm}, \mathrm{E}=$ ?
(c) $r=18 \mathrm{~cm}, \mathrm{E}=$ ?
(a) For a point inside the spherical conductor, electric field, $\mathrm{E}=$ zero
(b) For point just on the surface of the conductor, E
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{R}^2} \\
&=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2}=\frac{9 \times 1.6 \times 10^2}{144 \times 10^{-4}} \\
&=10^5 \mathrm{NC}^{-1}
\end{aligned}
$$
$=10^5 \mathrm{NC}^{-1}$. For point outside the conductor,
(c) For point outside the conductor,
$$
\begin{aligned}
\mathrm{E} &=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}=\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} \\
&=\frac{9 \times 1.6 \times 10^2}{324 \times 10^{-4}}=4.4 \times 10^4 \mathrm{NC}^{-1} .
\end{aligned}
$$
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