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:

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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. To maximize the force, we need to maximize the product $|q_1 q_2|$. If the initial charge is $Q$ and $q$ is removed, then one object has charge $q$ and the other has charge $Q-q$. The product of the charges is $q(Q-q)$. To find the maximum value, we can take the derivative with respect to $q$ and set it to zero: $\frac{d}{dq}[q(Q-q)] = Q - 2q = 0$, solving for $q$ gives $q = Q/2$.

A Small Object Has Charge Q