Verified Solution • 2025–26 Session moderate

QAIPMT-2009-Ch-2091: NEET 2010 Physics - Class 12 AIPMT-2009-Ch-2091 Solution: Two positive ions, each carrying a charge q, are separated by a …

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)

Multiple Choice Options

  • A 4πϵ0Fd2e2\frac{4\pi\epsilon_0 Fd^2}{e^2}
  • B 4πϵ0Fe2d2\sqrt{\frac{4\pi\epsilon_0 Fe^2}{d^2}}
  • C 4πϵ0Fd2e2\sqrt{\frac{4\pi\epsilon_0 Fd^2}{e^2}}
  • D 4πϵ0Fd2q2\frac{4\pi\epsilon_0 Fd^2}{q^2}

Step-by-Step Verified Solution

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 = 1 4 � _ 0 ( q ) ( q ) d 2 F = q 2 4 � _ 0 d 2 q 2 = 4 � _ 0 F d 2 q = 4 � _ 0 F d 2    ...(i) Since, q = ne where, n = number of electrons missing from each ion e = magnitude of charge on electron _ n = q e n = 4 � _ 0 F d 2 e    (Using (i)) = 4 � _ 0 F d 2 e 2

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