Two equal point charges ' $q$ ' each exert a force ' $F$ ' on each other, when they are placed distance '…

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
  1. $\frac{1}{\sqrt{\mathrm{k}}}$
  2. $\sqrt{\mathrm{k}}$
  3. $\frac{\sqrt{\mathrm{k}}}{2}$
  4. $\frac{2}{\sqrt{\mathrm{k}}}$

Solution

Force between the two charges in air, $\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} \cdot \mathrm{q}}{\mathrm{x}^2}$
Force between the two charges when in a medium of dielectric constant k, $\begin{aligned} & \mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0 \mathrm{k}} \cdot \frac{\mathrm{q} \cdot \mathrm{q}}{\mathrm{y}^2} \\ & \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{x}^2} \times \frac{4 \pi \varepsilon_0 \mathrm{k} \cdot \mathrm{y}^2}{\mathrm{q}^2} \end{aligned}$ $\begin{array}{ll}\quad 1 & =k \frac{y^2}{x^2} \quad \ldots\left(\because F_1=F_2\right) \\ \therefore \quad & \frac{y}{x}=\frac{1}{\sqrt{k}}\end{array}$

Asked in: MHT CET 2024 (03 May Shift 1)

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