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# what is the potential on this surface? ?

For a single, isolated point charge carrying a charge of pC, q=57.7 pC, one equipotential surface consists of a sphere of radius r1=29.4 mm centered on the point charge.

What is the potential on this surface?

To draw an additional equipotential surface separated by V 11.7 V from the previous surface, how far from the point charge should this second surface be? This surface must also meet the condition of being farther from the point charge than the original equipotential surface is.

### 1 Answer

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- NCSLv 72 months agoFavorite Answer
potential = kQ / r

where k = 1/4πε₀ = 8.99e9N·m²/C²

so

potential = 8.99e9N·m²/C² * 57.7e-12C / 0.0294m

potential = 17.6 V ◄

then

11.7 V = 8.99e9N·m²/C² * 57.7e-12C / r

solves to

r = 0.0443 m = 44.3 mm ◄

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