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# A metal sphere of radius 30 cm has a net charge of$3.0 \times 10 ^ { - 8 } C$. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere 's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 500 V?

Solution

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Given: $R=30\,\,\rm cm$, $Q=3\cdot10^{-8}\,\,\rm C$

a)

We can calculate electric field with equation:

\begin{align*} E&=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\\ E&=\frac{1}{4\pi8.85\cdot10^{-12}}\frac{3\cdot10^{-8}}{0.3^2}\\ \end{align*}

$\boxed{E=3\,\,\rm kN/C}$

b)

Electric potential is equal to:

\begin{align*} V&=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\\ V&=\frac{1}{4\pi8.85\cdot10^{-12}}\frac{3\cdot10^{-8}}{0.3^2}\\ \end{align*}

$\boxed{V=0.9\,\,\rm V}$

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