What is the percentage change in distance if the force of attraction between two point charges increases to 4 times keeping magnitude of charges constant ?

Answer

To solve the problem, we need to analyze how the force of attraction between two point charges changes when the distance between them is altered. Let's break down the steps:

### Step-by-Step Solution:

1. **Understand the Formula for Electrostatic Force**:
The electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's Law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

2. **Initial Force**:
Let the initial force be \( F \):
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

3. **New Force**:
According to the problem, the new force \( F' \) is 4 times the initial force:
\[ F' = 4F = 4 \left( \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \right) \]

4. **Relate New Force to New Distance**:
The new force can also be expressed in terms of the new distance \( r' \):
\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(r')^2} \]

5. **Set the Equations Equal**:
Since both expressions represent the new force \( F' \), we can set them equal to each other:
\[ 4 \left( \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \right) = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(r')^2} \]

6. **Cancel Common Terms**:
Cancel \( \frac{1}{4 \pi \epsilon_0} q_1 q_2 \) from both sides:
\[ 4 \cdot \frac{1}{r^2} = \frac{1}{(r')^2} \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ (r')^2 = \frac{r^2}{4} \]

8. **Taking the Square Root**:
Taking the square root of both sides results in:
\[ r' = \frac{r}{2} \]

9. **Calculate the Change in Distance**:
The change in distance is:
\[ \Delta r = r - r' = r - \frac{r}{2} = \frac{r}{2} \]

10. **Calculate the Percentage Change**:
The percentage change in distance is given by:
\[ \text{Percentage Change} = \left( \frac{\Delta r}{r} \right) \times 100 = \left( \frac{\frac{r}{2}}{r} \right) \times 100 = 50\% \]

### Final Answer:
The percentage change in distance is **50%**.