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 touched with sphere $C$ is touched with sphere $A$ and then it is placed midway between $A$ and $B$. Then the magnitude of Net force on $C$ is

• A. $1\times {10}^{-5}N$
• B. $1.5\times {10}^{-5}N$
• C. $2\times {10}^{-5}N$
• D. $3\times {10}^{-5}N$

Answer 1

The correct option is D $3\times {10}^{-5}N$

Let these two identical charge spheres have charge $q$ and placed at a distance $r$ apart from each other. Then,
$F=\frac{K{q}_1{q}_2}{r^2}$
$or,3\times {10}^{-5}=\frac{K{q}^2}{r^2}$ [Since $q_1=q_2=q$] ..........(1)
Now another identical uncharged sphere is touched with A, hence charge will be distributed equally.
Hence, charge of A = charge on C=$\frac{q}{2}$
Force on C due to A
$F_1=\frac{K.\frac{q}{2}.\frac{q}{2}}{{\left(\frac{r}{2}\right)}^2}$
$=\frac{K{q}^2}{r^2}$
Force on C due to B
$F_2=\frac{K.\frac{q}{2}.q}{{\left(\frac{r}{2}\right)}^2}$
$=2\frac{K{q}^2}{r^2}$
As both are opposite, net force =$F_2-F_1=2\frac{K{q}^2}{r^2}-\frac{K{q}^2}{r^2}=\frac{K{q}^2}{r^2}=3\times {10}^{-5}$N [Using (1)]