Question

the table below shows four pairs of atoms and the electric charge of each atom. the atoms in each pair are located the same distance from each other. which pair of atoms is experiencing the strongest repulsive force? a. pair s b. pair p c. pair r d. pair q

Explanation:

Step1: Recall Coulomb's law

The magnitude of the electrostatic force between two charged particles is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is a constant, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them. Since $r$ is the same for all pairs, we only need to consider the product $|q_1q_2|$.

Step2: Calculate product for Pair P

For Pair P with $q_1 = + 1$ and $q_2=-5$, $|q_1q_2|=|1\times(- 5)| = 5$.

Step3: Calculate product for Pair Q

For Pair Q with $q_1 = + 2$ and $q_2=-6$, $|q_1q_2|=|2\times(-6)| = 12$.

Step4: Calculate product for Pair R

For Pair R with $q_1 = + 3$ and $q_2 = + 4$, $|q_1q_2|=|3\times4|=12$.

Step5: Calculate product for Pair S

For Pair S with $q_1 = + 4$ and $q_2 = + 2$, $|q_1q_2|=|4\times2| = 8$.

Step6: Compare products for repulsive - force pairs

We are looking for the repulsive - force pair (same - sign charges). Among the pairs with same - sign charges (Pair R and Pair S), the product $|q_1q_2|$ is larger for Pair R ($12$) than for Pair S ($8$).

Answer:

C. Pair R