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Physics Question 28 – NEET-UG 2025

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Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :

When identical conducting spheres touch, the total charge is shared equally between them.

● Video Walkthrough
Step 1: Initial Force and Charges✦ Active

Initially, sphere A has charge qA=q and sphere B has charge qB=q. The force of repulsion between them is given by Coulomb's Law:

F=kqAqBr2=kq2r2
Step 2: Charge Redistribution on Spheres A and B○ Expand

1. A third identical uncharged sphere C (charge qC=0) touches sphere A (charge q). Since they are identical, the total charge is shared equally. The new charge on A is qA′=q+02=q2. Sphere C also acquires a charge of q2.

2. Sphere C (now with charge q2) then touches sphere B (charge q). Again, the total charge is shared equally. The new charge on B is qB′=q+q22=3q22=3q4.

💡 Teacher's Secret Hint

Remember that charge is conserved and distributes equally among identical conductors in contact.

Step 3: Calculate New Force of Repulsion○ Expand

After the process, sphere A has charge qA′=q2 and sphere B has charge qB′=3q4. The distance r remains the same. The new force of repulsion F′ is:

F′=kqA′qB′r2=k(q2)(3q4)r2=k3q28r2

Substitute the initial force F=kq2r2 into the expression for F′:

F′=38(kq2r2)=38F
💡 Teacher's Secret Hint

Ensure you use the final charges on spheres A and B for the new force calculation.

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