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Question

Physics Question on Electrostatic potential

On moving a charge of 20C20\, C by 2cm,2J2\, cm, 2\, J of work is done, then the potential difference between the points is

A

0.1V0.1\,V

B

8V8\,V

C

2V2\,V

D

0.5V0.5\,V

Answer

0.1V0.1\,V

Explanation

Solution

Potential difference between two points in an electric field
VAVB=Wq0V_{A}-V_{B} =\frac{W}{q_{0}}
where WW is work done by moving charge q0q_{0} from point AA to BB
Here, W=2J,q0=20cW=2J, q_{0}=20\,c
VAVB=220V_{A}-V_{B}=\frac{2}{20}
=0.1=0.1 volt