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?
\begin{tabular}{|c|c|c|}hline pair & atom x & atom y \hline p & +1 & -5 \hline q & +2 & -6 \hline r & +3 & +4 \hline s & +4 & +2 \hlineend{tabular}
a. pair s
b. pair q
c. pair r
d. pair p

Explanation:

Step1: Recall Coulomb's law

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

Step2: Calculate product for Pair P

For Pair P with $q_1 = 1$ and $q_2=5$, $|q_1q_2|=|1\times5| = 5$.

Step3: Calculate product for Pair Q

For Pair Q with $q_1 = 2$ and $q_2 = 6$, $|q_1q_2|=|2\times6|=12$.

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

Pairs Q and R have the largest product of charge magnitudes among the four pairs. Since we are looking for repulsive force (charges must have the same - sign), Pair R has two positive - charged atoms, while Pair Q has opposite - charged atoms (which attract).

Answer:

C. Pair R