Two small similar metal spheres $\mathrm{A}$ and $\mathrm{B}$ having charges $4 \mathrm{q}$ and $-4…
Two small similar metal spheres $\mathrm{A}$ and $\mathrm{B}$ having charges $4 \mathrm{q}$ and $-4 \mathrm{q}$, when placed at a certain distance apart, exert an electric force $\mathrm{F}$ on each other. When another identical uncharged sphere $\mathrm{C}$, first touched with $\mathrm{A}$ then with $\mathrm{B}$ and then removed to infinity, the force of interaction between $\mathrm{A}$ and $\mathrm{B}$ for the same separation will be $\frac{F}{x}$, then find the value of $x ?$
7
8
9
10
Solution
$\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(4 \mathrm{q})(-4 \mathrm{q})}{\mathrm{r}^{2}}$
when $\mathrm{C}$ is touched with $\mathrm{A}$, then charge on $\mathrm{A} \& \mathrm{C}$ each $=$ $2 \mathrm{q}$ after that $\mathrm{C}$ is touched with $\mathrm{B}$, charge on $\mathrm{B}=\frac{2 \mathrm{q}+(-4 \mathrm{q})}{2}=-\mathrm{q}$
Now, force $\mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{q})(-\mathrm{q})}{\mathrm{r}^{2}} \Rightarrow \mathrm{F}^{\prime}=\frac{\mathrm{F}}{8}$