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 \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

### Step 1: Understand the initial conditions
We know that the initial force \( F \) between the two charges is 100 N. We denote the initial distance between the charges as \( d \).

### Step 2: Increase the distance by 50%
If the distance between the charges is increased by 50%, the new distance \( d' \) can be calculated as:

\[ d' = d + 0.5d = 1.5d \]

### Step 3: Calculate the new force
Using Coulomb's law again, the new force \( F' \) when the distance is \( d' \) is given by:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(d')^2} \]

Substituting \( d' = 1.5d \):

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(1.5d)^2} \]

### Step 4: Simplify the expression
We can simplify \( (1.5d)^2 \):

\[ (1.5d)^2 = 2.25d^2 \]

Thus, the new force becomes:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{2.25d^2} \]

### Step 5: Relate the new force to the initial force
We know that the initial force \( F \) is:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} \]

Now we can express \( F' \) in terms of \( F \):

\[ F' = \frac{F}{2.25} \]

### Step 6: Substitute the initial force value
Substituting \( F = 100 \, \text{N} \):

\[ F' = \frac{100}{2.25} \approx 44.44 \, \text{N} \]

### Conclusion
The new force between the two charges after increasing the distance by 50% is approximately \( 44.44 \, \text{N} \).

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