Question

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

Answer 1

Answer on Question #69267 Physics / Other

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

Solution:

The force between charges $Q - q$ and $q$ is equal

$$F = k \frac{(Q - q)q}{r^2} = k \frac{Qq - q^2}{r^2}.$$

The maximum force condition (extremum of $F$) is given by

$$\frac{dF}{dq} = 0.$$

Thus

$$\frac{d}{dq}(Qq - q^2) = Q - 2q = 0.$$

Finally

$$q = \frac{Q}{2}.$$

**Answers**: $q = \frac{Q}{2}$.

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