Two identical bodies in which charges are `40muC` and `-20muC`. They are some distance apart. Now they are touched and kept at the same distance. The ratio of the initial to the final force between them is

Options

A. `8:1`
B. `4:1`
C. `1:8`
D. `1:1`

Answer

To solve the problem, we need to find the ratio of the initial force to the final force between two charged bodies after they are touched and then separated. Let's break this down step by step.

### Step 1: Calculate the Initial Force
The initial force \( F_{\text{initial}} \) between two charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by Coulomb's law:

\[ F_{\text{initial}} = k \frac{|Q_1 \cdot Q_2|}{r^2} \]

Where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( Q_1 = 40 \, \mu C = 40 \times 10^{-6} \, C \)
- \( Q_2 = -20 \, \mu C = -20 \times 10^{-6} \, C \)

Substituting the values:

\[ F_{\text{initial}} = k \frac{|40 \times 10^{-6} \cdot (-20) \times 10^{-6}|}{r^2} \]

Calculating the product of the charges:

\[ |Q_1 \cdot Q_2| = |40 \times (-20)| \times 10^{-12} = 800 \times 10^{-12} = 8 \times 10^{-10} \]

Thus,

\[ F_{\text{initial}} = k \frac{8 \times 10^{-10}}{r^2} \]

### Step 2: Calculate the Final Charge After Touching
When the two bodies are touched, the total charge is shared equally between them. The total charge \( Q_{\text{total}} \) is:

\[ Q_{\text{total}} = Q_1 + Q_2 = 40 \, \mu C + (-20 \, \mu C) = 20 \, \mu C \]

Since they are identical, the final charge on each body \( Q_f \) will be:

\[ Q_f = \frac{Q_{\text{total}}}{2} = \frac{20 \, \mu C}{2} = 10 \, \mu C = 10 \times 10^{-6} \, C \]

### Step 3: Calculate the Final Force
The final force \( F_{\text{final}} \) between the two bodies after they are charged and separated again is:

\[ F_{\text{final}} = k \frac{|Q_f \cdot Q_f|}{r^2} \]

Substituting the final charge:

\[ F_{\text{final}} = k \frac{|10 \times 10^{-6} \cdot 10 \times 10^{-6}|}{r^2} \]

Calculating the product of the final charges:

\[ |Q_f \cdot Q_f| = |10 \times 10^{-6} \cdot 10 \times 10^{-6}| = 100 \times 10^{-12} = 1 \times 10^{-10} \]

Thus,

\[ F_{\text{final}} = k \frac{1 \times 10^{-10}}{r^2} \]

### Step 4: Calculate the Ratio of Initial to Final Force
Now we can find the ratio of the initial force to the final force:

\[ \text{Ratio} = \frac{F_{\text{initial}}}{F_{\text{final}}} = \frac{k \frac{8 \times 10^{-10}}{r^2}}{k \frac{1 \times 10^{-10}}{r^2}} = \frac{8 \times 10^{-10}}{1 \times 10^{-10}} = 8 \]

Thus, the ratio of the initial force to the final force is:

\[ \text{Ratio} = 8:1 \]

### Conclusion
The ratio of the initial force to the final force between the two bodies is \( 8:1 \).