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Problem 5: Two positive point charges of 15 × 10-10C and 13 × 10-10C are placed 12 cm apart. Find the work done in bringing the two charges 4 cm closer.


Since the charges have same sign, therefore, the force between them is repulsive. The force will increase as they come closer to one another. So, we have to find the average force and then apply the formula for work done, that is, W = F.d
Force between the charges when they are 12 cm apart.

When they are brought 4 cm closer, their distance is 12 – 4 = 8 cm, and hence force of repulsion at the final position

The average force acting on the charges when they are brought 4 cm close is (F1 +F2)/2. Therefore,

Now work done in moving the charge = Fav.d, therefore,

This is the answer.


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