Physics Circle home Student Login

Home›Chapters›Electrostatics PYQs›Electric Charge & Coulomb's Law›Question 50

MHT CET 2021 (23 Sep, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from MHT CET 2021 (23 Sep, Shift 2)

Q50MHT CET 2021 · 23 SEP · SHIFT 2Easy

Electric Charge & Coulomb's Law

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 from each ion will be (ee = charge on electron,ε0\text{electron}, \varepsilon_0 = permittivity of free space)

Show answer and solution

Answer: (d)

  1. Each ion has charge q=neq = ne, where nn = electrons missing.
  2. F=14πε0n2e2d2⇒n2=4πε0Fd2e2F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{n^2e^2}{d^2} \Rightarrow n^2 = \dfrac{4\pi\varepsilon_0 F d^2}{e^2}
  3. So the answer is (d) n=4πε0Fd2e2n = \sqrt{\dfrac{4\pi\varepsilon_0 F d^2}{e^2}}.
Report a mistake

More questions on Electric Charge & Coulomb's Law