two charges placed at a distance experience a force of XN. what is the force between the charges, when each charge is doubled, the distance between them halved and the charge placed in a medium of di electric constant 2.52.
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Answer:
2.52XN
Explanation:
initial force=XN=K×q1q2/r^2
new Force=2.52K×(2q1)(2q2)/(r/2)^2=2.52XN
here K=9×10^9
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