Electrostatics Question 479

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 (VBVA) is then [Haryana CEE 1996]

Options:

A) + 2kV

B) ? 2 kV

C) + 200 V

D) ? 200 V

Show Answer

Answer:

Correct Answer: A

Solution:

Work done W=Q(VBVA)(VBVA)=WQ

=10×1035×106J/C=2kV



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