Question 829: Electrostats & Capacitance

Question

Two equal point charges ' $q$ ' each exert a force ' $F$ ' on each other, when they are placed distance ' $x$ ' apart in air. When the same charges are placed distance ' $y$ ' apart in a medium of dielectric constant ' $k$ ', they exert the same force. The ratio of distance ' $y$ ' to ' $x$ ' is equal to

Accepted Answer

When two equal point charges, each of magnitude $ q $, exert a force $ F $ on each other when placed distance $ x $ apart in air, the force is given by:
$ F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2} $
Here, $\varepsilon_0$ is the permittivity of free space.
When these charges are placed distance $ y $ apart in a medium with a dielectric constant $ k $, they exert the same force $ F $:
$ F_2 = \frac{1}{4 \pi \varepsilon_0 k} \frac{q^2}{y^2} $
Since the forces $ F_1 $ and $ F_2 $ are equal, we have:
$ \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2} = \frac{1}{4 \pi \varepsilon_0 k} \frac{q^2}{y^2} $
Simplifying the equation:
$ k \frac{y^2}{x^2} = 1 $
From this, we determine the ratio of distance $ y $ to $ x $:
$ \frac{y}{x} = \frac{1}{\sqrt{k}} $
Thus, the ratio of distance $ y $ to $ x $ in the medium is $\frac{1}{\sqrt{k}}$.

Answer

$\frac{1}{\sqrt{\mathrm{k}}}$

Answer

$\sqrt{\mathrm{k}}$

Answer

$\frac{\sqrt{\mathrm{k}}}{2}$

Answer

$\frac{2}{\sqrt{\mathrm{k}}}$

Electrostats & Capacitance

Two equal point charges ' q ' each exert a force ' F ' on each other, when they are placed distance ' x ' apart in air. When the same charges are placed distance ' y ' apart in a medium of dielectric constant ' k ', they exert the same force. The ratio of distance ' y ' to ' x ' is equal to

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