Question

Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest under electrical forces. What should be the charge on C and where should it be clamped?

Answer 1

This question means that the net force on any one of the three charge will be zero.

In such case, we need to find distance.

Let the distance be x from the 2q charge and magnitude of the charge be Q.  

Therefore,

  Force by Charge A = kqQ/x²

  Force by charge B = kQ(2q)/(d - x)²

Now, it should be there,

       kqQ/x² = kQ(2q)/(d - x)²

        1/x² = 2/(d - x)²

        (1/x)² = [√2/(d - x)]²

1/x = √2/(d - x) or 1/x = -√2/(d - x)

d = x(√2 + 1) or d = -x(√2 - 1)

x = d/(√2 + 1) or x = -d/(√2 - 1)

This should be the two points where net force will be zero. So we can Place a third charge C at any of the one position.

Hope it helps.

Answer 2

Answer:

x = d/(√2 + 1)

Explanation: