Question

Two identical copper spheres $A$ and $B$ equally charged, repel each other with a force of $4\times {10}^{-5}N$. A third similar copper sphere $C$ carrying no charge is now brought into contact with $A$ and placed mid-way between $A$ and $B$.what force would $C$ experience? $(4\times {10}^{-5})N$
(Comment: charge on $A$ and $C$ is $q/2$ each and on $B$ it is $q$. It $r$ is the distance between $A$ and $B$, $q/r$ is $2\times {10}^{-7})$

Answer 1

in the first situation $F=\frac{k{q}^2}{r^2}=4\times {10}^{-5}N$

Now in the 2nd situation when identical sphere $C$ touched with sphere $A$ then there is transfer of charge from $A$ to $C$ till both have same potential , in this situation both will acquired equal charge this gives charge on both sphere will be $q/2$ (since total charge should be equal to $q$due to conservation of charge )

so, from the figure we can see that net force on sphere $C$ will be effective of both by $A$ and $B$

$F_C={\overrightarrow{F}}_{CA}+{\overrightarrow{F}}_{CB}$

$=\frac{k{q}^2/4}{r^2/4}-\frac{k\left(q/2\right)\left(q\right)}{r^2/4}$

$=\frac{k{q}^2}{r^2}(1-2)$

$=-\frac{k{q}^2}{r^2}=-4\times {10}^{-5}N$

this shows that force experienced by sphere C will be $4\times {10}^{-5}N$