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.

pair	atom x	atom y
q	-3	-5
r	-3	+4
s	+1	+2

which pair of atoms is experiencing the strongest attractive force?
a. pair s
b. pair r
c. pair p
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 the distance $r$ is the same for all pairs, we only need to consider the product of the magnitudes of the charges $|q_1q_2|$.

Step2: Calculate the product of charges for each pair

For pair P: $|+ 4|\times|-5|=20$.
For pair Q: $|-3|\times|-5| = 15$.
For pair R: $|-3|\times|+4|=12$.
For pair S: $|+1|\times|+2| = 2$.

Step3: Compare the products

The largest product is 20 for pair P.

Answer:

C. Pair P