A 10 μ C charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:

Solution

Physics sir · 3 steps

With r fixed, the force goes as the product of the two parts, and that product peaks when you split equally.

1. Write the force. Let one part be q, so the other is Q - q with Q = 10 μC. The distance is fixed at 1 cm, so only the product of the charges can change the force.

   F = 1/4πε₀ q(Q-q)/r²

2. Maximise the product. Set d/dq[q(Q-q)] = Q - 2q = 0, which gives q = Q/2. The second derivative is -2, so this is a maximum.

   Option	Parts (μC)	Product q(Q-q) (μC²)
   (a)	9 and 1	9
   (b)	5 and 5	25
   (c)	8 and 2	16
   (d)	7 and 3	21

3. So. The repulsion is greatest when both parts are 5 μC, which is option (b).

The idea to keep. When a fixed total is split into two parts, their product is largest for an equal split. So for maximum force between two parts of one charge, take q = Q/2 each.

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