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Electrostatics question

2010 · Q186
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Electrostatics question

2010 · Q186

NEETPhysicsElectrostaticsMCQ+4 / −1
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)
  1. A
    4πε0Fd2e2{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}e24πε0​Fd2​
  2. B
    4πε0Fe2d2\sqrt {{{4\pi {\varepsilon _0}F{e^2}} \over {{d^2}}}}d24πε0​Fe2​​
  3. C
    4πε0Fd2e2\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}}e24πε0​Fd2​​
  4. D
    4πε0Fd2q2{{4\pi {\varepsilon _0}F{d^2}} \over {{q^2}}}q24πε0​Fd2​
View written solutionFree

Correct answer: C

According to Coulomb’s law, the force of repulsion between the two positive ions each of charge q, separated by a distance d is given by

F=14πε0(q)(q)d2F = {1 \over {4\pi {\varepsilon _0}}}{{\left( q \right)\left( q \right)} \over {{d^2}}}F=4πε0​1​d2(q)(q)​

F=q24πε0d2F = {{{q^2}} \over {4\pi {\varepsilon _0}{d^2}}}F=4πε0​d2q2​

q2=4πε0Fd2{q^2} = 4\pi {\varepsilon _0}F{d^2}q2=4πε0​Fd2

q=4πε0Fd2q = \sqrt {4\pi {\varepsilon _0}F{d^2}} q=4πε0​Fd2​   ...(i)

Since, q = ne
where, n = number of electrons missing from each ion
e = magnitude of charge on electron

∴n=qe \therefore n = {q \over e}∴n=eq​

n=4πε0Fd2en = {{\sqrt {4\pi {\varepsilon _0}F{d^2}} } \over e}n=e4πε0​Fd2​​   (Using (i))

= 4πε0Fd2e2\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}} e24πε0​Fd2​​

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