A small object has charge Q.Charge q is removed from it and placed on a second small object.The two objects are placed 1 m apart.For the force that each object exerts on the other to be a maximum, q should be:

Question

Quizplus · Accepted Answer

The force between two charges is given by Coulomb's law, $F = k \frac{|q_1 q_2|}{r^2}$, where $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and $k$ is Coulomb's constant. For the force to be maximum, the product of $q_1$ and $q_2$ should be maximum. If the original charge is $Q$ and $q$ is removed, then one object has charge $q$ and the other $Q - q$. The product of the charges is $q(Q - q)$. To maximize this product, set the derivative with respect to $q$ equal to zero: $\frac{d}{dq}[q(Q - q)] = Q - 2q = 0$, solving this gives $q = Q/2$.

A Small Object Has Charge Q