Question

Two equally charged identical metallic spheres $A$ and $B$ repel each other with a force $2\times {10}^{-5}\text{N}$, when placed in air (neglect the dimension of sphere as they are very small). Another identical uncharged sphere $C$ is touched to $B$ and then placed at the mid point of line joining $A$ and $B$. What is the net electrostatic force on $C$ ?

• A. $1\times {10}^{-5}\text{N}$, toward $BA$
• B. $2\times {10}^{-5}\text{N}$, towards $AB$
• C. $4\times {10}^{-5}\text{N}$, towards $BA$
• D. $0.5\times {10}^{-5}\text{N}$, towards $AB$

Answer 1

The correct option is B $2\times {10}^{-5}\text{N}$, towards $AB$


Let the initial charge on the sphere $A$ and $B$ be $+q$ and separated by $r$. The force of repulsion,
$F=\frac{k{q}^2}{r^2}=2\times {10}^{-5}\text{N}$
When the sphere $C$ is touched with $B$; the charge of $B$ will get distributed equally on $B$ and $C$ due to their identical nature.
$\therefore q_B=q_C=\frac{q}{2}$


Force on sphere $C$ by $A$,
$F_{CA}=\frac{k{q}^2}{2{\left(\frac{r}{2}\right)}^2}=\frac{2k{q}^2}{r^2}=2F$
$F_{CA}$ is directed from $A$ to $B$

Similarly, force on the sphere $C$ by sphere $B$
$F_{CB}=\frac{k\left(\frac{q}{2}\right)\left(\frac{q}{2}\right)}{{\left(\frac{r}{2}\right)}^2}=\frac{k{q}^2}{r^2}$
$\Rightarrow F_{CB}=F$
and $F_{CB}$ is directed from $B$ to $A$.
Thus net force on $C$ is,
$F_c=2F-F=F$
$\Rightarrow F_c=2\times {10}^{-5}\text{N}$ from $A$ to $B$