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Two equipotential surfaces surround a +2.90 x 10-8-C point charge. How far is the 220-V surface from the 85.0-V surface?

Two equipotential surfaces surround a +2.90 x 10-8-C point charge. How far is the 220-V surface from the 85.0-V surface?
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Two equipotential surfaces surround a +2.90 x 10-8-C point charge. How far is the 220-V surface from the 85.0-V surface?

User Troi
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The potential difference isV=220−85=135.
Now the potential V at distance d from a point charge Q isV=Q/4πϵ0d
If you solve this for d and plug in the potential difference you will get the answer.
User Rambalachandran
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