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Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \;cm$. the charge on each is $6.5 \times 10^{-7}\; C?$ Suppose the spheres $A$ and $B$ have identical sizes.A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between $A$ and $B?$
$8.4 \times 10^{-2} \;N$.
$5.7 \times 10^{-3} \;N$.
$3.5 \times 10^{-5} \;N$.
$6.7 \times 10^{-2} \;N$.
Solution
Distance between the spheres, $A$ and $B, r=0.5\,m$
Initially, the charge on each sphere, $q=6.5 \times 10^{-7} \,C$
When sphere $A$ is touched with an uncharged sphere $C, q / 2$ amount of charge from $A$ will transfer to sphere $C$. Hence, charge on each of the spheres, $A$ and $C ,$ is $q / 2 .$
When sphere $C$ with charge $q / 2$ is brought in contact with sphere $B$ with charge $q$, total charges on the system will divide into two equal halves given as,
$\frac{1}{2}\left(q+\frac{q}{2}\right)=\frac{3 q}{4}$
Hence, charge on each of the spheres, $C$ and $B ,$ is $\frac{3 q}{4}$ Force of repulsion between sphere A having charge $q / 2$ and sphere $B$ having charge $\frac{3 q}{4}$ is $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\frac{q}{2} \times \frac{3 q}{4}}{r^{2}}$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 q^{2}}{8 r^{2}}=\frac{9 \times 10^{9} \times 3 \times\left(6.5 \times 10^{-7}\right)^{2}}{8 \times(0.5)^{2}}$$=5.703 \times 10^{-3}\, N$
Therefore, the force of attraction between the two spheres is $5.703 \times 10^{-3} \;N$.
