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, $F_{A B}=\frac{1}{4 \pi \epsilon _{0}}.\frac{q . q}{r^{2}}=\frac{1}{4 \pi \epsilon _{0}}.\frac{q^{2}}{r^{2}}$

Finally, force on $C$
$F_{C}=F_{C B}-F_{C A}=\frac{1}{4 \pi \left(\epsilon \right)_{0}}.\frac{\left(\frac{q}{2}\right) \left(q\right)}{\left(\frac{r}{2}\right)^{2}}-\frac{1}{4​\pi ε​​​ _{0}}.\frac{\left(\frac{q}{2}\right) \left(\frac{q}{2}\right)}{\left(\frac{r}{2}\right)^{2}}=\frac{1}{4 \pi ​ \left(ε​\right)_{0}}.\frac{q^{2}}{r^{2}}$
$So \, \, F_{C}=F_{A B} \, = \, F$