Question

coulombs law
used to calculate the force between two charged particles
f ∝ (\frac{q_{1}q_{2}}{r^{2}})
(q_{1}) = electrical charge on particle 1
(q_{2}) = electrical charge on particle 2
r = distance between them
force of attraction = positive and negative charges
force of repulsion = same charges
changing the magnitude of the charged particle
f ∝ (\frac{q_{1}q_{2}}{r^{2}})
if (q_{2}) becomes larger, f will be
changing particle distance
f ∝ (\frac{q_{1}q_{2}}{r^{2}})
if the distance between (q_{1}) and (q_{2}) becomes larger, f will be

Explanation:

Step1: Analyze force - charge relationship

From $F\propto\frac{q_1q_2}{r^{2}}$, when $r$ is constant, if $q_2$ becomes larger, since $F$ is directly proportional to the product of charges, $F$ will increase.

Step2: Analyze force - distance relationship

From $F\propto\frac{q_1q_2}{r^{2}}$, when $q_1$ and $q_2$ are constant, if $r$ becomes larger, the denominator $r^{2}$ becomes larger, and since $F$ is inversely - proportional to $r^{2}$, $F$ will decrease.

Answer:

If $q_2$ becomes larger, $F$ will be larger. If the distance between $q_1$ and $q_2$ becomes larger, $F$ will be smaller.