Question

Two small identical ball $P$ and $Q$, each of mass $0.866\text{g}$, carry identical charges and are suspended by two threads of equal length $50\text{cm}$ each from the same point. At equilibrium, they are separated by a distance of $50\text{cm}$. Find the charge on each ball (in the nearest integer in $\text{nC}$).
Take $g=10{\text{m/s}}^2$ and $\tan {60}^{\circ }=1.7$.

Answer 1

FBD of charged balls in equilibrium is shown in the figure.
Let us consider a ball, say $P$.
There are three forces acting on it:
$\left(i\right)$ Tension $\left(T\right)$ in the thread
$\left(ii\right)$ Force $\left(mg\right)$ of gravity
$\left(iii\right)$ Electric force $\left(F\right)$

Considering the equilibrium,
$T\sin {60}^{\circ }=mg$, and
$T\cos {60}^{\circ }=F=\frac{k{q}^2}{r^2}$

So, $\tan {60}^{\circ }=\frac{mg{r}^2}{k{q}^2}$
$\Rightarrow q=\sqrt{\frac{mg{r}^2}{k\tan {60}^{\circ }}}$
$\Rightarrow q=\sqrt{\frac{0.866\times {10}^{-3}\times 10\times (0.5{)}^2}{9\times {10}^9\times 1.7}}$
$\Rightarrow q=3.76\times {10}^{-7}=376\text{nC}$