Question 320: Electrostatics & Capacitance

Question

If two charges q$$_1$$ and q$$_2$$ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

Accepted Answer

$$
\begin{aligned}
& 	ext { dielectric constant }=\mathrm{K} \\
& F_{	ext {medium }}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}ight)} 	imes \frac{q_{1} q_{2}}{d^{2}} \\
& \because \quad F_{	ext {air }}=F_{	ext {medium }} \\
& \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{\left(d_{\mathrm{air}}ight)^{2}}=\frac{1}{4 \pi\left(\mathrm{K} \varepsilon_{0}ight)} \frac{q_{1} q_{2}}{d^{2}} \\
& 	herefore \quad d_{	ext {air }}=d\sqrt{\mathrm{K}}
\end{aligned}
$$

Answer

$$d\sqrt k$$

Answer

$$1\,.\,5d\sqrt k$$

Answer

$$k\sqrt d$$

Answer

$$2d\sqrt k$$

Electrostatics & Capacitance

If two charges q$_1 and q_2$ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

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