Question 870: Electrostats & Capacitance

Question

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 from each ion will be ($$\varepsilon$$ = charge on $$\varepsilon_k$$ = permittivity of free space)

Accepted Answer

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

Answer

$$
\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{~d}^2}{\mathrm{e}^2}}
$$

Answer

$$
\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}}{\mathrm{e}^2}}
$$

Answer

$$
\sqrt{\frac{4 \pi \varepsilon_0 F d^2}{\mathrm{e}}}
$$

Answer

$$
\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}
$$

Electrostats & Capacitance

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 from each ion will be ($\varepsilon = charge on \varepsilon_k$ = permittivity of free space)

Choose the correct answer: