A force \( F \) acts between two spheres A and B carrying charges \( +Q \) and \( -Q \), respectively, and placed distance ‘\( r \)’ apart. The force acting on a third sphere C having charge \( +Q \) placed at the mid-point of A and B will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8F \) towards B
Use Coulomb’s law and add the two forces on C as vectors.
\( \displaystyle F = \frac{kQ^{2}}{r^{2}} \)
C is at distance \( r/2 \) from each sphere, so each force has magnitude:
\( \displaystyle F' = \frac{kQ^{2}}{(r/2)^{2}} = 4\,\frac{kQ^{2}}{r^{2}} = 4F \)
Directions:
\( \text{A } (+Q) \text{ repels C } (+Q)\text{: towards B} \)
\( \text{B } (-Q) \text{ attracts C } (+Q)\text{: towards B} \)
Both forces point the same way, so they add:
\( F_{\mathrm{net}} = 4F + 4F = 8F \text{ towards B} \)