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Question

A metal sphere of radius 30 cm has a net charge of

3.0×108C3.0 \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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Answered 2 years ago
Answered 2 years ago
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Given: R=30cmR=30\,\,\rm cm, Q=3108CQ=3\cdot10^{-8}\,\,\rm C

a)

We can calculate electric field with equation:

E=14πϵ0QR2E=14π8.85101231080.32\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*}

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

b)

Electric potential is equal to:

V=14πϵ0QR2V=14π8.85101231080.32\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*}

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

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