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Bringing a charge qq from very far away to a positon close to another fixed charge...

Bringing a charge qq from very far away to a positon close to another fixed charge changes the potential energy of the two charges by ΔU=−2.0nJΔ⁢U=−2.0nJ If the potential difference for the first charge is ΔV=−2.0VΔ⁢V=−2.0V, what is its charge?

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Answer #1

ΔU = q(ΔV)

so, q = ΔU/(ΔV)

= (-2nJ)/(-2V)

= 1 nC - the charge

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