Two point charge `+2 C` and `+6 C` repel each other with a force of `12 N` . If a charge of `-4 C` is given to each other of these charges , the force will now be

Options

A. 4 N (repulsive)
B. 4 N (attractive)
C. 12 N (attractive)
D. 8 N (repulsive)

Answer

To solve the problem step by step, 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 \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

### Step 1: Identify the initial charges and force
We have two charges:
- \( Q_1 = +2 \, C \)
- \( Q_2 = +6 \, C \)

The initial force \( F \) between them is given as \( 12 \, N \).

### Step 2: Calculate the new charges after adding \(-4 \, C\)
When we add a charge of \(-4 \, C\) to each of these charges, the new charges become:
- \( Q_1' = Q_1 - 4 \, C = +2 \, C - 4 \, C = -2 \, C \)
- \( Q_2' = Q_2 - 4 \, C = +6 \, C - 4 \, C = +2 \, C \)

### Step 3: Calculate the new force using the new charges
The new force \( F' \) can be calculated using the modified charges:

\[ F' = k \frac{|Q_1' Q_2'|}{r^2} \]

Substituting the new charges:

\[ F' = k \frac{|-2 \times 2|}{r^2} = k \frac{4}{r^2} \]

### Step 4: Relate the new force to the initial force
From Coulomb's law, we know that:

\[ \frac{F}{F'} = \frac{Q_1 Q_2}{Q_1' Q_2'} \]

Substituting the known values:

\[ \frac{12}{F'} = \frac{(2)(6)}{(-2)(2)} \]

Calculating the right side:

\[ \frac{12}{F'} = \frac{12}{-4} \]

This simplifies to:

\[ \frac{12}{F'} = -3 \]

### Step 5: Solve for the new force \( F' \)
Cross-multiplying gives:

\[ 12 = -3 F' \]

Thus,

\[ F' = -4 \, N \]

### Step 6: Interpret the result
The negative sign indicates that the force is attractive.

### Final Answer
The new force \( F' \) is \( 4 \, N \) and it is attractive.

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