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Question

Physics Question on coulombs law

Two positive ions, each carrying a charge qq, are separated by a distance dd. If FF 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)

A

4πε0Fd2e2 \frac{ 4 \pi \varepsilon_0 F d^2 }{ e^2}

B

4πε0Fe2d2 \sqrt{ \frac{ 4 \pi \varepsilon_0 F e^2 }{ d^2}}

C

4πε0Fd2e2 \sqrt{ \frac{ 4 \pi \varepsilon_0 F d^2 }{ e^2}}

D

4πε0Fd2q2 \frac{ 4 \pi \varepsilon_0 F d^2 }{ q^2}

Answer

4πε0Fd2e2 \sqrt{ \frac{ 4 \pi \varepsilon_0 F d^2 }{ e^2}}

Explanation

Solution

According to Coulomb's law, the force of repulsion between the two positive ions each of charge qq, separated by a distance dd is given by
F=14πε0(q)(q)d2F = \frac{ 1}{ 4 \pi \varepsilon_0 } \frac{(q)(q)}{ d^2}
F=q24πε0d2F = \frac{ q^2}{ 4 \pi \varepsilon_0 d^2}
q2=4πε0Fd2q^2 = 4 \pi \varepsilon_0 F \, d^2
q=4πε0Fd2q = \sqrt{ 4 \pi \varepsilon_0 \, F \, d^2 } ..(i)
Since, q=neq = ne
where,
n=n = number of electrons missing from each ion
e=e = magnitude of charge on electron
n=qe\therefore n = \frac{ q}{ e}
n=4πε0Fd2en = \frac{ \sqrt{ 4 \pi \varepsilon_0 F \, d^2 }}{ e} (Using (i))
=4πε0Fd2e2= \sqrt{ \frac{ 4 \pi \varepsilon_0 F \, d^2 }{ e^2 }}