Question

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

Explanation:

Step1: Recall Coulomb's Law for Force Type

Coulombic repulsion occurs when two charges have the same sign (both positive or both negative), and attraction occurs when charges have opposite signs. We analyze the product of \( q_1 \) and \( q_2 \) for each set (since the sign of \( F_c \) indicates force type: positive for repulsion, negative for attraction, considering \( k>0 \), \( d^2>0 \)).

Step2: Analyze Set 1

For Set 1: \( q_1 = -1 \, \text{C} \), \( q_2 = +1 \, \text{C} \).
Calculate \( q_1 \times q_2 \):
\( (-1) \times (+1) = -1 \).
A negative product means the force is attractive (since \( F_c=\frac{k(q_1q_2)}{d^2} \), negative \( q_1q_2 \) gives negative \( F_c \), indicating attraction).

Step3: Analyze Set 2

For Set 2: \( q_1 = -1 \, \text{C} \), \( q_2 = -1 \, \text{C} \).
Calculate \( q_1 \times q_2 \):
\( (-1) \times (-1) = +1 \).
A positive product means the force is repulsive (since positive \( q_1q_2 \) gives positive \( F_c \), indicating repulsion, as like charges repel).

Answer:

Set 2