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 repul - Physics MCQ Physics | ExamTest.live | ExamTest.live

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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:

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

Solution & Step-by-step Explanation

1. Initial Force: F = \frac{k q ^{2}}{r ^{2}} .2. Contact with Sphere A: Third sphere C (uncharged, 0) touches A (charge q). Since they are identical, charge is shared equally. New charge on A: q_{A} = \frac{q + 0}{2} = \frac{q}{2} . New charge on C: q_{C} = \frac{q}{2} .3. Contact with Sphere B: Sphere C (charge q /2) touches B (charge q). New charge on B: q_{B} = \frac{q + q /2}{2} = \frac{3 q /4}{1} = \frac{3 q}{4} .4. Final Force: F^{'} = \frac{k q _{A} q _{B}}{r ^{2}} = \frac{k ( q /2 ) ( 3 q /4 )}{r ^{2}} = \frac{3}{8} \frac{k q ^{2}}{r ^{2}} = \frac{3 F}{8} .

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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:

A

\frac{3 F}{5}

B

\frac{2 F}{3}

C

\frac{F}{2}

D

\frac{3 F}{8}

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