A) \[\frac{3F}{4}\]
B) \[\frac{F}{2}\]
C) F
D) \[\frac{3F}{8}\]
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 \[F=\frac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[{{F}_{f}}\] will become \[3/{{8}^{th}}\]of \[{{F}_{i}}.\] Initially charges on both sphere, q \[F=\frac{K{{q}^{2}}}{{{d}^{2}}}\] .............(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 \[F'=\frac{k\times \frac{q}{2}\times \frac{3q}{4}}{{{d}^{2}}}\] \[F'=\frac{3F}{8}\]
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