Question
The work done in carrying a charge of 5μC from a point A to a point B in an electric field is 10mJ. The potential difference (VB−VA) is
Answer: Option A
:
A
Work done W=Q(VB−VA)⇒(VB−VA)=WQ=10×10−35×10−6J/C=2kV
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:
A
Work done W=Q(VB−VA)⇒(VB−VA)=WQ=10×10−35×10−6J/C=2kV
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