The force between two point charges in air is 100 N. If the distance between them is increased by 50%, then the force between two charges will be nearly equal to

Options

A. 50N
B. 56N
C. 100N
D. 44N

Answer

To solve the problem, we will use Coulomb's Law, which states that the force \( F \) between two point charges is given by:

\[ F = k \frac{q_1 q_2}{r^2} \]

where:
- \( F \) is the force between the charges,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step 1: Identify the initial conditions
We know that the initial force \( F \) is 100 N when the distance between the charges is \( r \).

\[ F = k \frac{q_1 q_2}{r^2} = 100 \, \text{N} \]

### Step 2: Determine the new distance
The problem states that the distance between the charges is increased by 50%. Therefore, the new distance \( r' \) can be calculated as:

\[ r' = r + 0.5r = 1.5r = \frac{3}{2}r \]

### Step 3: Calculate the new force
Using Coulomb's Law again for the new distance \( r' \):

\[ F' = k \frac{q_1 q_2}{(r')^2} = k \frac{q_1 q_2}{\left(\frac{3}{2}r\right)^2} \]

### Step 4: Simplify the expression for the new force
Substituting \( r' \) into the equation:

\[ F' = k \frac{q_1 q_2}{\left(\frac{9}{4}r^2\right)} = k \frac{q_1 q_2}{r^2} \cdot \frac{4}{9} \]

### Step 5: Substitute the initial force into the new force equation
From our initial condition, we know that \( k \frac{q_1 q_2}{r^2} = 100 \, \text{N} \). Therefore, we can substitute this into the equation for \( F' \):

\[ F' = \frac{4}{9} \cdot 100 \, \text{N} = \frac{400}{9} \, \text{N} \approx 44.44 \, \text{N} \]

### Conclusion
Thus, the new force between the two charges when the distance is increased by 50% is approximately 44.44 N.

### Final Answer
The force between the two charges will be nearly equal to **44.44 N**.

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