Correct option is (B) $10^{-10}$
Given : Gravitational force $=$ Electrostatics force
Let the distance between the rwo charges be r.
$$
\begin{array}{l}
\Rightarrow \frac{\mathrm{G}(\mathrm{m}) \mathrm{m}}{\mathrm{r}^{2}}=\frac{\mathrm{K} \mathrm{Q} \cdot \mathrm{Q}}{\mathrm{r}^{2}} \\
\Rightarrow \frac{\mathrm{G}}{\mathrm{K}}=\frac{\mathrm{Q}^{2}}{\mathrm{~m}^{2}} \\
\Rightarrow \sqrt{\frac{6.67 \times 10^{-11}}{9 \times 10^{9}}}=\frac{\mathrm{Q}}{\mathrm{m}}
\end{array}
$$
$\therefore \frac{\mathrm{Q}}{\mathrm{m}}$ is the order of $10^{-10}$
Two particles each of mass $\mathrm{m}$ and carrying charge $\mathrm{Q}$ are separated by some distance. If they are in equilibrium under mutual gravitational and electrostatic forces, then $Q / m$ (in $\mathrm{C} / \mathrm{Kg}$ ) is of the order of:
A $10^{-6}$
B $10^{-10}$
(C) $10^{-15}$
D $10^{-20}$
A $10^{-6}$
B $10^{-10}$
(C) $10^{-15}$
D $10^{-20}$
Two particles each of mass $\mathrm{m}$ and carrying charge $\mathrm{Q}$ are separated by some distance. If they are in equilibrium under mutual gravitational and electrostatic forces, then $Q / m$ (in $\mathrm{C} / \mathrm{Kg}$ ) is of the order of:
A $10^{-6}$
B $10^{-10}$
(C) $10^{-15}$
D $10^{-20}$
A $10^{-6}$
B $10^{-10}$
(C) $10^{-15}$
D $10^{-20}$