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

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

Solution

Two positive ions each carrying a charge $q$ are kept at a distance $\mathrm{d}$, then it is found that force of repulsion between them is $\begin{aligned} \mathrm{F} & =\frac{\mathrm{kqq}}{\mathrm{d}^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{q q}{\mathrm{~d}^2} \end{aligned}$ where $\begin{array}{ll} \text { where } & \mathrm{q}=\mathrm{ne} \\ \therefore & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{n}^2 \mathrm{e}^2}{\mathrm{~d}_2} \\ \Rightarrow & \mathrm{n}=\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}_2}{\mathrm{e}^2}} \end{array}$

Asked in: NEET 2010 (Screening)

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