charges of +3 μc and -5 μc are 2 mm from each other. the -5 μc charge is replaced with a +5 μc charge. how…

Explanation:

Step1: Recall Coulomb's law

The electrical force between two charges is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Analyze the original situation

Let $q_1 = + 3\ \mu C=3\times10^{- 6}\ C$, $q_2=-5\ \mu C = - 5\times10^{-6}\ C$, and $r = 2\ mm=2\times10^{-3}\ m$. The force $F_1=k\frac{q_1q_2}{r^2}=k\frac{(3\times10^{-6})\times(-5\times10^{-6})}{(2\times10^{-3})^2}$, and the negative sign indicates the nature of the force (attractive).

Step3: Analyze the new - situation

When $q_2$ is replaced with $+5\ \mu C = + 5\times10^{-6}\ C$, the new force $F_2=k\frac{q_1q_2}{r^2}=k\frac{(3\times10^{-6})\times(5\times10^{-6})}{(2\times10^{-3})^2}$.

Step4: Compare the magnitudes and directions

The magnitudes of $F_1$ and $F_2$ are equal since $|F_1|=|F_2| = k\frac{|3\times10^{-6}|\times|5\times10^{-6}|}{(2\times10^{-3})^2}$. The original force was attractive (opposite - sign charges), and the new force is repulsive (same - sign charges), so the force has the same magnitude but opposite direction.

Answer:

same force, but in the opposite direction