Two positive ions, each carrying a charge ' $\mathrm{q}$ ' are separated by a distance ' $\mathrm{d}$ '. If…

Two positive ions, each carrying a charge ' $\mathrm{q}$ ' are separated by a distance ' $\mathrm{d}$ '. If ' $\mathrm{F}$ ' is the force of repulsion between the ions, the number of electrons from each ion will be $(\mathrm{e}=$ charge on electron, $\varepsilon_0=$ permittivity of free space)
  1. $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{~d}^2}{\mathrm{e}^2}}$
  2. $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}}{\mathrm{e}^2}}$
  3. $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}}{\mathrm{e}}}$
  4. $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}$

Solution

$\begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{~d}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(\mathrm{ne})^2}{\mathrm{~d}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{n}^2 \mathrm{e}^2}{\mathrm{~d}^2} \\ & \therefore \mathrm{n}=\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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