Question 945: Electrostats & Capacitance

Question

Two positives 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)

Accepted Answer

According to CoulombÃ¢ÂÂs law, the force of repulsion between the two positive ions each of charge q, separated by a distance d is given by
$$F = {1 \over {4\pi {\varepsilon _0}}}{{\left( q ight)\left( q ight)} \over {{d^2}}}$$
$$F = {{{q^2}} \over {4\pi {\varepsilon _0}{d^2}}}$$
$${q^2} = 4\pi {\varepsilon _0}F{d^2}$$
$$q = \sqrt {4\pi {\varepsilon _0}F{d^2}} $$   ...(i)
Since, q = ne
where, n = number of electrons missing from each ion
e = magnitude of charge on electron
$$ 	herefore n = {q \over e}$$
$$n = {{\sqrt {4\pi {\varepsilon _0}F{d^2}} } \over e}$$   (Using (i))
= $$\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}} $$

Answer

$${{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}$$

Answer

$$\sqrt {{{4\pi {\varepsilon _0}F{e^2}} \over {{d^2}}}} $$

Answer

$$\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}} $$

Answer

$${{4\pi {\varepsilon _0}F{d^2}} \over {{q^2}}}$$

Electrostats & Capacitance

Two positives 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)

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