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In a Van De Graaff generator, a spherical metal shell isto be $15 \times 10^6 \mathrm{~V}$ electrode. The dielectric strength of the gas surrounding the electrode is $5 \times 10^{-7} \mathrm{~V} / \mathrm{m}$. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
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Given, $\mathrm{V}=15 \times 10^6 \mathrm{~V}$
Dielectric strength of the surrounding gas, $\mathrm{E}=5 \times 10^7 \mathrm{~V} \mathrm{~m}^{-1}$
Radius of spherical shell, $r=$ ?
For a spherical shell, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}$
If $\sigma$ be surface density of charge, then
$\mathrm{q}=4 \pi \mathrm{r}^2 \sigma \quad \therefore \quad \mathrm{V}=\frac{4 \pi \mathrm{r}^2 \sigma}{4 \pi \varepsilon_0 \mathrm{r}}=\mathrm{r} \frac{\sigma}{\varepsilon_0}=\mathrm{rE}$
${\left[\because \mathrm{E}=\frac{\sigma}{\varepsilon_0}\right]$ we have, $\mathrm{r}=\mathrm{V} / \mathrm{E} }$
$=\frac{15 \times 10^6}{5 \times 10^7}=0.3 \mathrm{~m}=30 \mathrm{~cm}$
Dielectric strength of the surrounding gas, $\mathrm{E}=5 \times 10^7 \mathrm{~V} \mathrm{~m}^{-1}$
Radius of spherical shell, $r=$ ?
For a spherical shell, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}$
If $\sigma$ be surface density of charge, then
$\mathrm{q}=4 \pi \mathrm{r}^2 \sigma \quad \therefore \quad \mathrm{V}=\frac{4 \pi \mathrm{r}^2 \sigma}{4 \pi \varepsilon_0 \mathrm{r}}=\mathrm{r} \frac{\sigma}{\varepsilon_0}=\mathrm{rE}$
${\left[\because \mathrm{E}=\frac{\sigma}{\varepsilon_0}\right]$ we have, $\mathrm{r}=\mathrm{V} / \mathrm{E} }$
$=\frac{15 \times 10^6}{5 \times 10^7}=0.3 \mathrm{~m}=30 \mathrm{~cm}$
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