There are two charge `+1mu c` and `+2muc` kept at certain separartion ,the ratio of electrostatic forces acting on them will be in the ratio

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Explanation

To find the ratio of the electrostatic forces acting on two charges \( Q_1 \) and \( Q_2 \), we can use Coulomb's Law. Let's denote the charges as follows:

- \( Q_1 = +1 \mu C = 1 \times 10^{-6} C \)
- \( Q_2 = +2 \mu C = 2 \times 10^{-6} C \)

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Let \( Q_1 = 1 \mu C \) and \( Q_2 = 2 \mu C \).
- Let the distance between the charges be \( R \).

2. **Apply Coulomb's Law**:
- According to Coulomb's Law, the force \( F \) between two point charges is given by:
\[ F = k \frac{|Q_1 Q_2|}{R^2} \]
where \( k \) is Coulomb's constant (\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)).

3. **Calculate the Force on Each Charge**:
- The force acting on \( Q_1 \) due to \( Q_2 \) (denoted as \( F_{12} \)):
\[ F_{12} = k \frac{|Q_1 Q_2|}{R^2} = k \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{R^2} \]
- The force acting on \( Q_2 \) due to \( Q_1 \) (denoted as \( F_{21} \)):
\[ F_{21} = k \frac{|Q_2 Q_1|}{R^2} = k \frac{(2 \times 10^{-6})(1 \times 10^{-6})}{R^2} \]

4. **Determine the Ratio of Forces**:
- Since \( F_{12} = F_{21} \), we can express the ratio of the forces:
\[ \frac{F_{12}}{F_{21}} = \frac{k \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{R^2}}{k \frac{(2 \times 10^{-6})(1 \times 10^{-6})}{R^2}} = 1 \]

5. **Conclusion**:
- The ratio of the electrostatic forces acting on the two charges is:
\[ \frac{F_{12}}{F_{21}} = 1 \]

### Final Answer:
The ratio of the electrostatic forces acting on the charges \( Q_1 \) and \( Q_2 \) is \( 1:1 \).

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