Question

two objects have charges of 6 c and 6 c and are 12 cm apart.
change a: increase one charge to 18 c
change b: decrease the distance to 6 cm
which change causes a greater increase in force?
support your answer with reasoning.

Explanation:

Step1: Recall Coulomb's Law

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

Step2: Calculate initial force

Given $q_1=6\ \text{C}$, $q_2=6\ \text{C}$, $r=0.12\ \text{m}$:
$$F_{\text{initial}} = k\frac{(6)(6)}{(0.12)^2} = k\frac{36}{0.0144} = 2500k$$

Step3: Calculate force after Change A

$q_1=18\ \text{C}$, $q_2=6\ \text{C}$, $r=0.12\ \text{m}$:
$$F_A = k\frac{(18)(6)}{(0.12)^2} = k\frac{108}{0.0144} = 7500k$$
Force increase: $\Delta F_A = 7500k - 2500k = 5000k$

Step4: Calculate force after Change B

$q_1=6\ \text{C}$, $q_2=6\ \text{C}$, $r=0.06\ \text{m}$:
$$F_B = k\frac{(6)(6)}{(0.06)^2} = k\frac{36}{0.0036} = 10000k$$
Force increase: $\Delta F_B = 10000k - 2500k = 7500k$

Step5: Compare force increases

$\Delta F_B > \Delta F_A$, so Change B causes a larger increase.

Answer:

Change B (decreasing the distance to 6 cm) causes a greater increase in force. The initial electrostatic force is $2500k$. After Change A, the force becomes $7500k$ (a $5000k$ increase); after Change B, the force becomes $10000k$ (a $7500k$ increase), which is a larger magnitude of increase.