Two particles , each of mass `m` and carrying charge `Q` , are separated by some distance. If they are in equilibrium under mutual gravitational and electrostatic force then `Q//m (in C//kg)` is of the order of

Options

A. `10^(-5)`
B. `10^(-10)`
C. `10^(-15)`
D. `10^(-20)`

Explanation

To solve the problem, we need to find the ratio \( \frac{Q}{m} \) when two particles of mass \( m \) and charge \( Q \) are in equilibrium under the influence of gravitational and electrostatic forces.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The electrostatic force \( F_e \) between the two charges is given by Coulomb's law:
\[ F_e = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{d^2} \]
- The gravitational force \( F_g \) between the two masses is given by Newton's law of gravitation:
\[ F_g = G \frac{m^2}{d^2} \]

2. **Set the Forces Equal**:
Since the particles are in equilibrium, the electrostatic force must be equal to the gravitational force:
\[ F_e = F_g \]
Therefore, we have:
\[ \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{d^2} = G \frac{m^2}{d^2} \]

3. **Cancel \( d^2 \)**:
The \( d^2 \) terms cancel out from both sides:
\[ \frac{1}{4 \pi \epsilon_0} Q^2 = G m^2 \]

4. **Rearrange the Equation**:
Rearranging the equation to find \( \frac{Q^2}{m^2} \):
\[ Q^2 = 4 \pi \epsilon_0 G m^2 \]

5. **Take the Square Root**:
Taking the square root of both sides gives:
\[ \frac{Q}{m} = \sqrt{4 \pi \epsilon_0 G} \]

6. **Substitute Values**:
Now, substituting the known values:
- \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( \epsilon_0 \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

Thus,
\[ \frac{Q}{m} = \sqrt{4 \pi (9 \times 10^9) (6.67 \times 10^{-11})} \]

7. **Calculate the Numerical Value**:
First, calculate \( 4 \pi \):
\[ 4 \pi \approx 12.57 \]
Now calculate:
\[ 4 \pi \epsilon_0 G \approx 12.57 \times 9 \times 10^9 \times 6.67 \times 10^{-11} \]
Simplifying gives:
\[ \approx 12.57 \times 6.67 \times 9 \times 10^{-2} \approx 10^{-10} \]

8. **Final Result**:
Therefore, the order of \( \frac{Q}{m} \) is approximately:
\[ \frac{Q}{m} \approx 10^{-10} \, \text{C/kg} \]

### Conclusion:
The ratio \( \frac{Q}{m} \) is of the order of \( 10^{-10} \, \text{C/kg} \).