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Two positive ions, each carrying a charge q, are separated by a distance d. If 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)

Option: 1

\frac{4\pi\varepsilon_{0}\text{Fd}^{2}}{\text{q}^{2}}


Option: 2

\frac{4\pi\varepsilon_{0}\text{Fd}^{2}}{\text{e}^{2}}


Option: 3

\sqrt{\frac{4\pi\varepsilon_{0}\text{Fe}^{2}}{\text{d}^{2}}}


Option: 4

\sqrt{\frac{4\pi\varepsilon_{0}\text{Fd}^{2}}{\text{e}^{2}}}


Answers (1)

best_answer

Two positive ions each carrying a charge q  are  kept at a distance d,
it is found that the force of repulsion between them is
\begin{aligned} & F=\frac{k q q}{\mathrm{~d}^2} \\ & =\frac{1}{4 \pi \mathrm{\varepsilon }_0} \frac{\mathrm{qq}}{\mathrm{d}^2} \end{aligned}
 
where \mathrm{q}= ne 
\begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \mathrm{\varepsilon }_0} \frac{\mathrm{n}^2 \mathrm{e}^2}{\mathrm{~d}^2} \\ & \mathrm{n}=\sqrt{\frac{4 \pi \mathrm{\varepsilon }_0 \mathrm{Fd}^2}{\mathrm{e}^2}} \end{aligned}

Posted by

himanshu.meshram

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