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 $C$,
$F_{A C}=\frac{k \frac{q}{2} \times q / 2}{(r / 2)^{2}}$
Similarly, between $B$ and $C$
$F_{B C}=\frac{k q \times q / 2}{(r / 2)^{2}}$
The magnitude of net force on $C, F^{\prime}=F_{B C}-F_{A C}$
Also $F=\frac{k q \times q}{r^{2}}=3 \times 10^{-5} \,N $
So, $ \frac{F}{F_{B C}-F_{A C}}=\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_{B C}-F_{A C}=F^{'} $
$\therefore F^{'}=3 \times 10^{-5}\, N$