QuestionAugust 1, 2025

When the distance between two charges is halved, the electrical force between them halves. doubles. is reduced by 1/4 quadruples. none of the above

When the distance between two charges is halved, the electrical force between them halves. doubles. is reduced by 1/4 quadruples. none of the above
When the distance between two charges is halved, the electrical force between them halves. doubles. is reduced by 1/4 quadruples. none of the above

Solution
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Answer

quadruples. Explanation 1. Identify the relevant formula The electrical force between two 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 charges, and r is the distance between them. 2. Analyze the effect of halving the distance If the distance r is halved, then the new distance is \frac{r}{2}. Substitute this into the formula: F' = k \frac{q_1 q_2}{(\frac{r}{2})^2} = k \frac{q_1 q_2}{\frac{r^2}{4}} = 4 \cdot k \frac{q_1 q_2}{r^2} = 4F. 3. Determine the change in force The force quadruples when the distance is halved.

Explanation

1. Identify the relevant formula<br /> The electrical force between two 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 charges, and $r$ is the distance between them.<br />2. Analyze the effect of halving the distance<br /> If the distance $r$ is halved, then the new distance is $\frac{r}{2}$. Substitute this into the formula: $F' = k \frac{q_1 q_2}{(\frac{r}{2})^2} = k \frac{q_1 q_2}{\frac{r^2}{4}} = 4 \cdot k \frac{q_1 q_2}{r^2} = 4F$.<br />3. Determine the change in force<br /> The force quadruples when the distance is halved.
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