There are two charges `+1 mu C ` and`-5 mu C` under mutual coulombic interaction . The ratio of magnitude of forces acting on themwill be `

Options

A. 1`:5`
B. `1:1`
C. `5:1`
D. `1:25`

Answer

To solve the problem of finding the ratio of the magnitudes of forces acting on two charges \( +1 \, \mu C \) and \( -5 \, \mu C \) under mutual Coulombic interaction, we can follow these steps:

### Step 1: Identify the Charges
Let:
- \( q_1 = +1 \, \mu C = +1 \times 10^{-6} \, C \)
- \( q_2 = -5 \, \mu C = -5 \times 10^{-6} \, C \)

### Step 2: Write the Expression for the Forces
According to Coulomb's law, the force between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant, \( |q_1 q_2| \) is the product of the magnitudes of the charges, and \( r \) is the distance between the charges.

### Step 3: Calculate the Forces
1. **Force on charge \( q_1 \) due to \( q_2 \)**:
\[ F_{12} = k \frac{|q_1 q_2|}{r^2} = k \frac{|(+1 \times 10^{-6})(-5 \times 10^{-6})|}{r^2} \]
\[ F_{12} = k \frac{5 \times 10^{-12}}{r^2} \]

2. **Force on charge \( q_2 \) due to \( q_1 \)**:
\[ F_{21} = k \frac{|q_1 q_2|}{r^2} = k \frac{|(-5 \times 10^{-6})(+1 \times 10^{-6})|}{r^2} \]
\[ F_{21} = k \frac{5 \times 10^{-12}}{r^2} \]

### Step 4: Compare the Forces
From the calculations above, we see that:
\[ F_{12} = F_{21} = k \frac{5 \times 10^{-12}}{r^2} \]
Thus, the magnitudes of the forces are equal.

### Step 5: Calculate the Ratio of Forces
The ratio of the magnitudes of the forces acting on the charges is:
\[ \frac{F_{12}}{F_{21}} = \frac{k \frac{5 \times 10^{-12}}{r^2}}{k \frac{5 \times 10^{-12}}{r^2}} = 1 \]

### Final Answer
The ratio of the magnitudes of the forces acting on the two charges is:
\[ \text{Ratio} = 1 : 1 \]