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A conducting sphere of radius $10 \mathrm{~cm}$ has an unknown charge. If the electric field $20 \mathrm{~cm}$ from the centre of the sphere is $1.5 \times 10^3 \mathrm{~N} / \mathrm{C}$ and points radially inward, what is the net charge on the sphere?
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Verified Answer
Given, $\mathrm{E}=-1.5 \times 10^3 \mathrm{~N} \mathrm{C}^{-1}(\mathrm{E}$ is taken negative because
it points inwardly opposite to $\mathrm{r}$ )
$$
\mathrm{r}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{q}=\text { ? }
$$
For outside point, the charge on the conducting sphere behaves as a point charge situated at the centre of the sphere.
By formula, $E=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}$
or $\quad \mathrm{q}=\mathrm{E} \times 4 \pi \varepsilon_0 \mathrm{r}^2$
$$
\begin{aligned}
&=-\frac{1.5 \times 10^3 \times(0.2)^2}{9 \times 10^9}=-\frac{6}{9} \times 10^{-9} \\
&=-6.67 \times 10^{-8} \mathrm{C} .
\end{aligned}
$$
it points inwardly opposite to $\mathrm{r}$ )
$$
\mathrm{r}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{q}=\text { ? }
$$
For outside point, the charge on the conducting sphere behaves as a point charge situated at the centre of the sphere.
By formula, $E=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}$
or $\quad \mathrm{q}=\mathrm{E} \times 4 \pi \varepsilon_0 \mathrm{r}^2$
$$
\begin{aligned}
&=-\frac{1.5 \times 10^3 \times(0.2)^2}{9 \times 10^9}=-\frac{6}{9} \times 10^{-9} \\
&=-6.67 \times 10^{-8} \mathrm{C} .
\end{aligned}
$$
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