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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Electrostatics

NEET 2010 · Physics · Electrostatics

Two positive ions, each carrying a charge q , are separated by a...

Two positive ions, each carrying a charge qq, are separated by a distance dd. If F\mathrm{F} is the force of repulsion between the ions, the number of electrons missing from each ion will be ( ee being the charge on an electron)

Options

  1. A4πε0Fd2e2\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}
  2. B4πε0Fe2 d2\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fe}^2}{\mathrm{~d}^2}}
  3. C4πε0Fd2e2\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}
  4. D4πε0Fd2q2\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{q}^2}

Correct answer

C. 4πε0Fd2e2\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}

Step-by-step solution

Two positive ions each carrying a charge qq are kept at a distance d\mathrm{d}, then it is found that force of repulsion between them is F=kqqd2=14πε0qqd2\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  where q=ne∴F=14πε0n2e2d2⇒n=4πε0Fd2e2\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}
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