Question

Two charged identical metal spheres $A$ and $B$ repel each other with a force of $3\times 10^{-5}N$ Another identical uncharged sphere $C$ is Louched wilh sphere $A$ and then il is placed mid-way between $A$ and $B$. Then, the magnitude of net force on $C$ is

• A. $1\times 10^{-5}N$
• B. $3\times 10^{-5}N$
• C. $2\times 10^{-5}N$
• D. $5\times 10^{-5}N$

Answer 1

Answer: (B)

Spheres $A$ and $B$ have the same charge of same nature. When uncharged sphere $C$ is touched with sphere $A$, then charge is transferred to $C$ and this charge is equally divided between two spheres.

Let initially the charge on spheres $A$ and $B$ is $q$.
Initially, the repelling force between charged spheres $A$ and $B=3\times 10^{-5}N$
When, sphere $C$, is placed between charged spheres $A$ and $B$, then
the force between $A$ and $C$,
$F_{AC}=\frac{k\frac{q}{2}\times q/2}{(r/2{)}^2}$
Similarly, between $B$ and $C$
$F_{BC}=\frac{kq\times q/2}{(r/2{)}^2}$
The magnitude of net force on $C,F^{\prime }=F_{BC}-F_{AC}$
Also $F=\frac{kq\times q}{r^2}=3\times 10^{-5}N$
So, $\frac{F}{F_{BC}-F_{AC}}=\frac{\left(k{q}^2/r^2\right)}{k\left[\frac{q\times q/2}{(r/2{)}^2}-\frac{q/2\times q/2}{(r/2{)}^2}\right]}$
Here, $F=3\times 10^{-5}N,F_{BC}-F_{AC}=F^{{}^{\prime }}$
$\therefore F^{{}^{\prime }}=3\times 10^{-5}N$