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\( 4 \times 10^{10} \) electrons are removed from a neutral metal sphere of diameter \( 20 \mathrm{~cm} \) placed in air.
The magnitude of the electric field (in \( N C^{-1} \) ) at a distance of \( 20 \mathrm{~cm} \) from its centre is:
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The magnitude of the electric field (in \( N C^{-1} \) ) at a distance of \( 20 \mathrm{~cm} \) from its centre is:
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Verified Answer
The correct answer is:
\( 1440 \)
Given, number of electrons removed $=4 \times 10^{10}$; diameter of metal sphere $=20 \mathrm{~cm}$; distance $\mathrm{r}=20 \mathrm{~cm}$. Therefore,
Electric field $=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}$
where $\frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9}$
$\Rightarrow q=\left(4 \times 10^{10}\right) \times\left(1.6 \times 10^{-19}\right)=6.4 \times 10^{-9}$
Electric field
$=\frac{9 \times 10^{9} \times 6.4 \times 10^{-9}}{\left(20 \times 10^{-2}\right)^{2}}=1440 \mathrm{~N} \mathrm{C}^{-1}$
Electric field $=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}$
where $\frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9}$
$\Rightarrow q=\left(4 \times 10^{10}\right) \times\left(1.6 \times 10^{-19}\right)=6.4 \times 10^{-9}$
Electric field
$=\frac{9 \times 10^{9} \times 6.4 \times 10^{-9}}{\left(20 \times 10^{-2}\right)^{2}}=1440 \mathrm{~N} \mathrm{C}^{-1}$
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