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Consider a thin spherical shell of radius $R$ consisting of uniform surface charge density $s$. The electric field at a point of distance $x$ from its centre and outside the shell is
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Verified Answer
The correct answer is:
inversely proportional to $x^{2}$
Electric field due to thin spherical shell at a distance $x$ outside from the centre is given as
$$
E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{x^{2}}
$$
where,
$$
\begin{aligned}
Q &=s \cdot A \\
&=s \times 4 \pi R^{2}
\end{aligned}
$$
$\therefore$ From Eq. (i), we get
$$
E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{s \times 4 \pi R^{2}}{x^{2}}=\frac{s R^{2}}{\varepsilon_{0} x^{2}}
$$
$$
\therefore \quad E \propto \frac{1}{x^{2}}
$$
$$
E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{x^{2}}
$$
where,
$$
\begin{aligned}
Q &=s \cdot A \\
&=s \times 4 \pi R^{2}
\end{aligned}
$$
$\therefore$ From Eq. (i), we get
$$
E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{s \times 4 \pi R^{2}}{x^{2}}=\frac{s R^{2}}{\varepsilon_{0} x^{2}}
$$
$$
\therefore \quad E \propto \frac{1}{x^{2}}
$$
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