Question 383: Electrostats & Capacitance

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 constant $$K=5$$ at '$$r / 5$$' $$\mathrm{cm}$$ apart will be:

Accepted Answer

In air $$F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}$$
In medium $$\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}$$

Answer

$$5 F$$

Answer

$$25 F$$

Answer

$$F / 5$$

Answer

$$F / 25$$

Electrostats & Capacitance

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

Choose the correct answer: