Question

Two equally charged small metal balls placed at a fixed distance experience a force $F$ . A similar uncharged metal 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. $2F$
• D. $4F$

Answer 1

Answer: (B)

Initially, $F_{AB}=\frac{1}{4\pi {ϵ}_0}.\frac{q.q}{r^2}=\frac{1}{4\pi {ϵ}_0}.\frac{q^2}{r^2}$

Finally, force on $C$
$F_C=F_{CB}-F_{CA}=\frac{1}{4\pi {\left(ϵ\right)}_0}.\frac{\left(\frac{q}{2}\right)\left(q\right)}{{\left(\frac{r}{2}\right)}^2}-\frac{1}{4\pi \epsilon {}_0}.\frac{\left(\frac{q}{2}\right)\left(\frac{q}{2}\right)}{{\left(\frac{r}{2}\right)}^2}=\frac{1}{4\pi {\left(\epsilon \right)}_0}.\frac{q^2}{r^2}$
$So{F}_C=F_{AB}=F$