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Electrostatics question

2025 · Q152
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Electrostatics question

2025 · Q152

NEETPhysicsElectrostaticsMCQ+4 / −1

Two identical charged conducting spheres AAA and BBB have their centres separated by a certain distance. Charge on each sphere is qqq and the force of repulsion between them is FFF. A third identical uncharged conducting sphere is brought in contact with sphere AAA first and then with BBB and finally removed from both. New force of repulsion between spheres AAA and BBB (Radii of AAA and BBB 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:

  1. A
    F2\frac{F}{2}2F​
  2. B
    3F8\frac{3 F}{8}83F​
  3. C
    3F5\frac{3 F}{5}53F​
  4. D
    2F3\frac{2 F}{3}32F​
View written solutionFree

Correct answer: B

NEET 2025 Physics - Electrostatics Question 1 English Explanation

$$\begin{aligned} F^{\prime} \& =\frac{\frac{K q}{2} \frac{3 q}{4}}{r^2} \\ F^{\prime} \& =\frac{3 F}{8} \end{aligned}$$

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