Two identically charged metallic spheres A and B placed at certain distance in air repel each other with a force F. If an uncharged identical metallic sphere C is touched with ‘A’ and removed, the new force between A and B is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( F/2 \)
When two identical conducting spheres touch, the charge is shared equally between them. The force between A and B follows Coulomb’s law: at a fixed distance it is proportional to the product of the charges.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q \times q}{r^{2}} \)
The uncharged sphere C takes half of the charge of A; B is untouched.
\( q_A = q/2 \)
\( q_B = q \)
\( \displaystyle F' = \frac{1}{4\pi\varepsilon_0}\,\frac{(q/2) \times q}{r^{2}} = \frac{F}{2} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Coulomb’s Law – “… Let the original charge on sphere A be q and that on B be q′. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\frac{qq'}{r^2} \)
neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2.”