Question

Force between two point charges \( q_1 \) and \( q_2 \) placed in vacuum at \( r \) cm apart is \( F \). Force between them when placed in a medium having dielectric \( K = 5 \) at \( r/5 \) cm apart will be:

• A. \( \frac{F}{25} \)
• B. \( 5F \)
• C. \( \frac{F}{5} \)
• D. \( 25F \)

Answer 1

The Correct Option is B

The force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) in a vacuum is given by Coulomb’s law:

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

where \(\epsilon_0\) is the permittivity of free space.

In a medium with dielectric constant \(K\), the permittivity changes from \(\epsilon_0\) to \(K \epsilon_0\). This reduces the effective force between the charges by a factor of \(K\). Thus, the force in the medium, if the distance remained \(r\), would be:

\[ F_{\text{medium}} = \frac{1}{4 \pi K \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{F}{K}. \]

For \(K = 5\), this becomes:

\[ F_{\text{medium}} = \frac{F}{5}. \]

Now, since the distance between the charges is reduced to \(\frac{r}{5}\), we need to adjust for this change. Coulomb’s force varies inversely with the square of the distance, so reducing the distance by a factor of \(\frac{1}{5}\) increases the force by a factor of \(\left(\frac{1}{5}\right)^{-2} = 25\).

Combining both effects (the dielectric and the reduced distance), the modified force \(F'\) in the medium is:

\[ F' = \frac{F}{5} \times 25 = 5F. \]

Thus, the force between the charges in the medium, with the distance changed to \(\frac{r}{5}\), is increased by a factor of 5 compared to the original force in a vacuum. Therefore, the answer is: \[ 5F. \]