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

Answer: (B)

$5F$

Answer 2

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: $5F.$