Question

As shown in the adjoining figure two charged particles each having charge $q$ and mass $m$ are $d$ distance apart from each other. If two particles are in equilibrium under the gravitational and electrostatic force, then the ratio $\frac{q}{m}\left(\text{C}/\text{kg}\right)$ is of the order

• A. ${10}^{-6}$
• B. ${10}^{-10}$
• C. ${10}^{-4}$
• D. ${10}^6$

Answer 1

The correct option is B ${10}^{-10}$

For equilibrium , electrostatic force $\left(F_E\right)$ is equal to the gravitational force $\left(F_G\right)$,

$F_E=F_G$


We know that,

$F_E=\frac{k{q}^2}{d^2}$

$F_G=\frac{G{m}^2}{d^2}$

Now, substituting the values in equilibrium equation,

$\frac{k{q}^2}{d^2}=\frac{G{m}^2}{d^2}$

$\Rightarrow \frac{1}{4\pi {ϵ}_0}\frac{q^2}{d^2}=\frac{G{m}^2}{d^2}$

$\Rightarrow \frac{q^2}{m^2}=G\times 4\pi {ϵ}_0$

$\Rightarrow \frac{q}{m}=\frac{\sqrt{G\times 4\pi {ϵ}_0}}{1}$

taking , $G=6.67\times {10}^{-11}$ and

$4\pi {ϵ}_0=\frac{1}{9\times {10}^9}$, we get,

$\therefore \frac{q}{m}=\sqrt{\frac{6.67\times {10}^{-11}}{9\times {10}^9}}\approx {10}^{-10}$

Hence, option (b) is the correct answer.