Two particles A and B having charges 20μC and -5μC respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?

Solution

Physics sir · 4 steps

Unlike charges, so the balance point lies outside the pair, on the side of the smaller charge.

1. Rule out between. Put a positive test charge between A and B. A pushes it towards B and B pulls it towards B, so both forces add and cannot cancel. A negative test charge has both forces reversed, and they still add.
2. Pick the side. Outside the pair the two forces point opposite ways. To balance, the test charge must sit closer to the smaller charge, -5 μC, so it goes to the right of B.

   (k (20 μC) q)/(5+x)² = (k (5 μC) q)/x²

3. Solve for x. Here x is the distance from B in cm. Taking square roots, 2/(5+x) = 1/x, so 2x = 5 + x and x = 5 cm. The charge q cancels, so this works for any third charge.
4. So. The third charge feels no net force 5 cm to the right of the -5 μC charge, that is 10 cm from A. That is option (b).

The idea to keep. For two unlike charges the null point is outside them, beside the smaller one. For like charges it is between them, closer to the smaller one. Then equate q₁/r₁² = q₂/r₂².

Answer from the NTA final answer key. Solution by PrepWiser.