Q. Two equally charged small balls placed at a fixed distance experience a force F. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is
A) $\frac{F}{2}$
B) $F$
C) $2\ F$
D) $4\ F$
Explanation
(b) Initially, $F_{AB} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q.q}{r^{2}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}$
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Finally, force on
$F_{C} = F_{AB} - F_{CA} = \frac{1}{4\pi\varepsilon_{0}}.\frac{\left( \frac{q}{2} \right)(q)}{\left( \frac{r}{2} \right)^{2}} - \frac{1}{4\pi\varepsilon_{0}}.\frac{\left( \frac{q}{2} \right)\left( \frac{q}{2} \right)}{\left( \frac{r}{2} \right)^{2}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}$
$\Rightarrow \ \ {\ \ F}_{C} = F_{AB}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.