Two point charges, $Q _ { 1 }$ and $Q _ { 2 }$, are separated by a distance $R$. If the magnitudes of both charges are halved and their separation is also halved, what happens to the electrical force that each charge exerts on the other one?

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Quizplus · Accepted Answer

The electrical force between two point charges is given by Coulomb's law, $F = k \frac{|Q_1 Q_2|}{R^2}$, where $k$ is Coulomb's constant, $Q_1$ and $Q_2$ are the magnitudes of the charges, and $R$ is the distance between them. If both charges and the distance are halved, the numerator of the fraction becomes $\frac{1}{4}$ of its original value ($\frac{1}{2} \times \frac{1}{2}$) and the denominator becomes $\frac{1}{4}$ of its original value ($\left(\frac{1}{2}\right)^2$). These changes cancel each other out, so the force remains the same.

Two Point Charges Q1Q _ { 1 }Q1​ And Q2Q _ { 2 }Q2​

Two Point Charges $Q _ { 1 }$ And $Q _ { 2 }$