Question

Two identical particles each of mass $M$ and charge $Q$ are placed a certain distance part. If they are in equilibrium under mutual gravitational and electric force, then calculate the ratio of ​$\frac{Q}{M}$ in SI units.

Answer 1

Step 1: Determination of the electric force between the two particles

The electric force between the two particles can be expressed as follows:

$F_e=\frac{K{Q}^2}{R^2}$

Here, $F_e$ is the electric force, $K$ is the proportionality constant, $Q$ is the charge of the particle, and $R$ is the distance between the particles.

Step 2: Determination of the gravitational force between the two particles

$F_g=\frac{G{M}^2}{R^2}$

Here, $F_g$ is the gravitational force, $G$ is the gravitational constant, and $M$ is the mass of the particles.

Step 3: Equilibrium of gravitational force and electrical force

Since the gravitational force and the electrical force are in equilibrium, it can be written as follows:

$$\begin{gathered} F_e=F_g \\ \frac{K{Q}^2}{R^2}=\frac{G{M}^2}{R^2} \\ \frac{Q^2}{M^2}=\frac{G}{K} \\ \frac{Q}{M}={\left(\frac{6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2}{9\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2}\right)}^{\frac{1}{2}} \\ \frac{Q}{M}=0.86\times {10}^{-10}\mathrm{C}/\mathrm{kg} \end{gathered}$$

Hence, the ratio of ​$\frac{Q}{M}$ in SI units is $0.86\times {10}^{-10}\mathrm{C}/\mathrm{kg}$.