JEE Main On 16 April 2018 Question 26
Question: Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere C is uncharged. Sphere C is first touched to A then to B, and the removed. As a result, the force between A and B would be equal to [JEE Main 16-4-2018]
Options:
A) $ \frac{3F}{4} $
B) $ \frac{F}{2} $
C) F
D) $ \frac{3F}{8} $
Show Answer
Answer:
Correct Answer: D
Solution:
The charge on A will be reduced to 3/4th of its original value,
& on B it will be halved. den as we know the formula for the electrostatic force between 2 charges
$ F=\frac{kq _1q _2}{r^{2}} $ $ F _{f} $ will become $ 3/8^{th} $ of
$ F _{i}. $ Initially charges on both sphere, q $ F=\frac{Kq^{2}}{d^{2}} $ ………….(1)
When sphere C will get touched with sphere A, then final charges on both will become $ \frac{q}{2}. $
Now, when this sphere C will get touched to sphere B, then final charges on both of them will be
$ q _{c}=qd=\frac{q/2+q}{2}=\frac{3q}{4} $ Now force between A and B will be
$ F’=\frac{k\times \frac{q}{2}\times \frac{3q}{4}}{d^{2}} $ $ F’=\frac{3F}{8} $
