The work done in carrying a charge of 5 μC from point A to a point B in an electric field is 10mJ. The potential difference VB- VA is then

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+2 KV.

-2 KV

+ 200 V

-200 V

answer is B.

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Detailed Solution

wAB=5μCVA−VB⇒10mJ5μC=VA−VB ∴ VA−VB=2kV or VB−VA=−2kV

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