NEET 2025 — Physics solution

NEET2025Physics

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 sphere 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:

A
$\frac{3 F}{5}$
B
$\frac{2 F}{3}$
C
$\frac{F}{2}$
D
$\frac{3 F}{8}$

Solution

Initially $F = \frac{k q ^{2}}{r ^{2}}$ .

Step 1: Third (uncharged) sphere touches A → charge equalises → A has $q /2$ , third sphere has $q /2$ .

Step 2: Third sphere ( $q /2$ ) touches B ( $q$ ) → total $3 q /2$ shared equally → B and third sphere each have $3 q /4$ .

Step 3: Third sphere removed. New charges: $q_{A} = q /2$ , $q_{B} = 3 q /4$ .

$F^{'} = \frac{k ( q /2 ) ( 3 q /4 )}{r ^{2}} = \frac{3 k q ^{2}}{8 r ^{2}} = \frac{3 F}{8}$ .

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