Coulomb's Law Multiple Choice Problems with detailed Solutions

2.1) A point charge 'Q' is to be divided in two parts 'q' and '(Q - q)'. To have the maximum electrostatic repulsion between the two parts (when they are placed at certain distance from each other), 'Q' and 'q' must hold the relation:
    A) q = Q/2
    B) q = Q/3
    C) q = 3Q/2
    D) q = Q/4
The electrostatic repulsion between the two point charge q and (Q - q), as governed by Coulomb's law, will have a strength F = kq(Q - q)/r² where r is the seperation between them (and it is fixed).
For the magnitude of the Coulomb force to be maximum, dF/dq = 0.
Therefore, option A  is the right choice. You may further verify that d²F/dq² is negative which implies that the force is maximum at q = Q/2.

2.2) Two point charges repel each other with a force of 100 N. If one of the charges is increased by 10% and the other is decreased by 10%, the new electrostatic force of repulsion at the same separation would be
    A) 100 N
    B) 121 N
    C) 99 N
    D) none of these



Let there be two point charges +Q and +q repelling each other with a force of 100 N, when kept at a distance 'r' from each other. By Coulomb's law, kQq/r² = 100. Let's assign it as equation number 1.
When Q is increased by 10% and q is decreased by 10%, the new electrostatic force is F' = kQ'q'/r² = k(Q+0.1Q)(q-0.1q)/r².
∴ The new electrostatic force of repulsion is 99N. Option C is the correct choice.

2.3) Two neutrons are kept at some distance from each other. They will
    A) attract each other
    B) repel each other
    C) neither attract nor repel
    D) cannot say

We know that neutrons are neutral particles i.e. they are chargeless. That's why two neutrons seperated by a certain distance will not interact electrostatically. However, though very weak, there will be a gravitational attraction between them by the virtue of their mass.

2.4) A negatively charged particle of mass m and charge -q starts moving around a fixed charge +Q in a circle of radius R. The time period of revolution T of the charge -q is given by
    A) T = (8Ï€³Îµᤱmr³/Qq)⁰·⁵
    B) T = (16Ï€³Îµᤱmr³/Qq)⁰·⁵
    C) T = (24Ï€³Îµᤱmr³/Qq)⁰·⁵
    D) None of these
The charged particle -q goes in a circular path of radius R around the fixed, positive charge +Q. The necessary centripetal force required to keep that particle in circular motion is provided by the electrostatic attraction between the two unlike charges.
 Also, as the seperation between the two charges is fixed (equal to r), the Coulomb force between them will be constant in magnitude. So the resulting circular motion will be a UCM (uniform circular motion). We have to obtain the period of this uniform circular motion. If the angular velocity is ω then,


Hindi Hearts

Previous Post Next Post