Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times where n is

Options

A. 5
B. 1
C. 0.16666666666667
D. 16

Explanation

To solve the problem step by step, 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:

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

where \( k \) is Coulomb's constant.

### Step 1: Define the Initial Conditions
Let the initial charges be \( Q_1 = Q \) and \( Q_2 = Q \). The initial distance between the charges is \( r \).

### Step 2: Calculate the Initial Force
Using Coulomb's Law, the initial force \( F \) between the charges is:

\[ F = k \frac{Q \cdot Q}{r^2} = k \frac{Q^2}{r^2} \]

### Step 3: Modify the Charges and Distance
According to the problem, each charge is doubled, so:

\[ Q_1' = 2Q \quad \text{and} \quad Q_2' = 2Q \]

The distance between the charges is halved, so:

\[ r' = \frac{r}{2} \]

### Step 4: Calculate the New Force
Now, we calculate the new force \( F' \) with the modified charges and distance:

\[ F' = k \frac{Q_1' \cdot Q_2'}{(r')^2} = k \frac{(2Q)(2Q)}{(\frac{r}{2})^2} \]

Calculating the denominator:

\[ (r')^2 = \left(\frac{r}{2}\right)^2 = \frac{r^2}{4} \]

Now substituting this back into the equation for \( F' \):

\[ F' = k \frac{(2Q)(2Q)}{\frac{r^2}{4}} = k \frac{4Q^2}{\frac{r^2}{4}} = k \frac{4Q^2 \cdot 4}{r^2} = k \frac{16Q^2}{r^2} \]

### Step 5: Relate the New Force to the Initial Force
Now we can relate the new force \( F' \) to the initial force \( F \):

\[ F' = 16 \left(k \frac{Q^2}{r^2}\right) = 16F \]

### Step 6: Determine the Value of \( n \)
According to the problem, the new force \( F' \) is \( n \) times the initial force \( F \):

\[ F' = nF \]

From our calculation, we found that \( F' = 16F \). Therefore, we can conclude that:

\[ n = 16 \]

### Final Answer
The value of \( n \) is \( 16 \).

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