Practice portal › Electric Charge and Properties › Electric Charge and Coulomb's Law

A charge of 4 μC is to be divided into two parts, the distance between the divided charges being constant. The magnitudes of the divided charges for which the force between them is maximum are

Asked in JEE Main 27th July 2nd Shift 2022 · Coulomb's law and superposition

Answer: (2) 2 μC and 2 μC

Step-by-step solution

With q and Q-q separated by a fixed r, F=(k q(Q-q))/(r²).

The force is largest when the product q(Q-q) is largest.

d/(dq)(Qq-q²)=Q-2q=0⇒ q=Q/2=2 μC.

So the charge splits equally: 2 μC and 2 μC.

Why the other options are wrong

More Electric Charge and Coulomb's Law questionsAll Electric Charge and Coulomb's Law questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer