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NEET PHYSICSMedium

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):

A

4\pi ε₀Fd² / e²

B

√(4\pi ε₀Fe² / d²)

C

√(4\pi ε₀Fd² / e²)

D

4\pi ε₀Fd² / q²

Step-by-Step Solution

According to Coulomb's Law, the electrostatic force of repulsion FF between two identical charges qq separated by a distance dd is given by: F=14πε0q2d2F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d^2}

Rearranging this equation to solve for the charge qq: q2=4πε0Fd2q^2 = 4\pi\varepsilon_0 F d^2 q=4πε0Fd2q = \sqrt{4\pi\varepsilon_0 F d^2}

According to the principle of quantization of charge, the charge qq on an ion is an integral multiple of the elementary charge ee (charge of an electron). If nn is the number of electrons missing from the ion, then: q=neq = ne

Substituting this into the expression for qq: ne=4πε0Fd2ne = \sqrt{4\pi\varepsilon_0 F d^2} n=4πε0Fd2en = \frac{\sqrt{4\pi\varepsilon_0 F d^2}}{e} n=4πε0Fd2e2n = \sqrt{\frac{4\pi\varepsilon_0 F d^2}{e^2}}

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