For a single, isolated point charge carrying a charge of q = 5.22 times 10^-11 C

Question

For a single, isolated point charge carrying a charge of q = 5.22 times 10^-11 C, one equipotential surface consists of a sphere of radius 0.0224 m centered on the point charge as shown. What is the potential on this surface? You would like to draw an additional equipotential surface, which is separated by 7.56 V from the previously mentioned surface. How far from the point charge should this surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface.

Explanation / Answer

here

Electric potential V = KQ/r

Q is charge

r is distance

K is constant = 9e9

so

V = (9e9 * 5.22 e -11/0.0224)

V = 20.97 Volts

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again V = KQ/(r+h)

(r+h) = 9e9 * 5.22e -11/(7.56+ 20.97)

r+h = 0.0164 m

h = 0.0165 - 0.0224

h = 0.006 m (ANSWER)

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