Question

unit exam-gravitational/electrostatic forces

	set 1	set 2
charge 1	-1 c	-1 c
charge 2	+1 c	-1 c
distance	3 × 10⁻⁸ m	3 × 10⁻⁸ m

which set has a coulombic attraction force?
remember : $f_c = \frac{k(q_1q_2)}{d^2}$
the sets exhibits an equivalent type of coulombic force.
set 1
set 2

Explanation:

Step1: Recall Coulomb's Law Sign Rule

Coulombic attraction occurs between opposite charges (one positive, one negative), repulsion between like charges (both positive or both negative). The sign of the force (from \( F_c = \frac{k(q_1q_2)}{d^2} \)) depends on \( q_1q_2 \): negative for attraction (opposite charges), positive for repulsion (like charges).

Step2: Analyze Set 1

For Set 1: \( q_1 = -1\ \text{C} \), \( q_2 = +1\ \text{C} \).
Calculate \( q_1q_2 = (-1)(+1) = -1 \).
Since \( q_1q_2 < 0 \), the force is attractive (negative sign indicates attraction in Coulomb's law context).

Step3: Analyze Set 2

For Set 2: \( q_1 = -1\ \text{C} \), \( q_2 = -1\ \text{C} \).
Calculate \( q_1q_2 = (-1)(-1) = +1 \).
Since \( q_1q_2 > 0 \), the force is repulsive (positive sign indicates repulsion).

Answer:

Set 1