Question

Two equally charged metal spheres $A$ and $B$ repel each other with a force of $4 \times 10^{-5}\, N$. Another identical uncharged sphere $C$ is touched to $A$ and then placed at the mid- point of the line joining the spheres $A$ and $B$. The net electric force on the sphere $C$ is

• A. $ 4 \times 10^{-5} N$ from C to A
• B. $ 4 \times 10^{-5} N$ from C to B
• C. $8 \times 10^{-5} N$ from C to A
• D. $8 \times 10^{-5} N$ from C to B

Answer 1

The Correct Option is A

Initially charge on $A$ and $B$ is $q$

Force, $F=\frac{k q^{2}}{d^{2}}=4 \times 10^{-5} N$ Now, $A$ is touched by $C$, then; Charge on $C=q / 2$ Charge on $A=q / 2$ So, force on $C= F _{A}+ F _{B}$ $=\frac{k q / 2 \cdot q / 2}{(d / 2)^{2}} \hat{ r }_{A C}+\frac{k q \cdot q / 2}{(d / 2)^{2}} \hat{ r }_{B C}$ $=\frac{k q^{2} / 4}{d^{2} / 4}-\frac{k q^{2} / 2}{d^{2} / 4}$ [as $\hat{ r }_{A C}=-\hat{ r }_{B C}$] $=\frac{k q^{2}}{d^{2}}(1-2)=-\frac{k q^{2}}{d^{2}}$ $=-4 \times 10^{-5} N$ So, force is of same magnitude but in opposite direction, i.e from $C$ to $A$ as suggested by minus sign.