A charge of `1 mu`C is divided into parts such that their charges are in the ratio of 2: 3. These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N) is

Options

A. 0.216
B. 0.00216
C. 0.0216
D. 2.16

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Divide the Charge
We have a total charge of \( Q = 1 \, \mu C \) (microcoulomb). The charge is divided in the ratio of 2:3.

Let the two parts of the charge be \( Q_1 \) and \( Q_2 \). According to the ratio:
\[ Q_1 : Q_2 = 2 : 3 \]

Let \( Q_1 = 2x \) and \( Q_2 = 3x \). Then:
\[ Q_1 + Q_2 = 1 \, \mu C \]
\[ 2x + 3x = 1 \, \mu C \]
\[ 5x = 1 \, \mu C \implies x = \frac{1}{5} \, \mu C \]

Now substituting back:
\[ Q_1 = 2x = 2 \times \frac{1}{5} \, \mu C = \frac{2}{5} \, \mu C \]
\[ Q_2 = 3x = 3 \times \frac{1}{5} \, \mu C = \frac{3}{5} \, \mu C \]

### Step 2: Use Coulomb's Law
Coulomb's law states that the electric force \( F \) between two charges is given by:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where:
- \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, N m^2/C^2 \)
- \( r \) is the distance between the charges, which is given as \( 1 \, m \).

### Step 3: Substitute the Values
Substituting the values of \( Q_1 \) and \( Q_2 \):
\[ F = 9 \times 10^9 \frac{\left(\frac{2}{5} \times 10^{-6}\right) \left(\frac{3}{5} \times 10^{-6}\right)}{1^2} \]

Calculating \( Q_1 Q_2 \):
\[ Q_1 Q_2 = \left(\frac{2}{5} \times 10^{-6}\right) \left(\frac{3}{5} \times 10^{-6}\right) = \frac{6}{25} \times 10^{-12} \]

Now substituting this back into the force equation:
\[ F = 9 \times 10^9 \frac{\frac{6}{25} \times 10^{-12}}{1} \]
\[ F = 9 \times 10^9 \times \frac{6}{25} \times 10^{-12} \]
\[ F = \frac{54}{25} \times 10^{-3} \, N \]

### Step 4: Calculate the Final Value
Calculating \( \frac{54}{25} \):
\[ \frac{54}{25} = 2.16 \]
Thus,
\[ F = 2.16 \times 10^{-3} \, N = 0.00216 \, N \]

### Final Answer
The electric force between the two charges is \( 0.00216 \, N \).

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