A 6.33 C charge is attracted by a force of 0.115 N to a second charge that is 1.44 m away. What is the value of the second charge? Include the + or - sign.
Answer (1)
To solve this problem, we need to use Coulomb's Law, which describes the force between two charges. The formula is given by:
F = k ⋅ r 2 ∣ q 1 ⋅ q 2 ∣
where:
F is the force between the charges (0.115 N in this case),
k is Coulomb's constant (approximately 8.99 × 1 0 9 N m 2 / C 2 ),
q 1 and q 2 are the magnitudes of the two charges (with q 1 = 6.33 C , and q 2 is what we're looking for),
r is the distance between the charges (1.44 m).
Rearranging the formula to solve for q 2 , we get:
q 2 = k ⋅ q 1 F ⋅ r 2
Substitute the given values:
q 2 = 8.99 × 1 0 9 N m 2 / C 2 ⋅ 6.33 C 0.115 N ⋅ ( 1.44 m ) 2
Calculate the result:
q 2 = 8.99 × 1 0 9 × 6.33 0.115 × 2.0736
q 2 ≈ 56.96727 × 1 0 9 0.238464
q 2 ≈ 4.18 × 1 0 − 12 C
Since the problem states that the charge is attracted, this means the second charge has the opposite sign to the first charge. Since the first charge is presumably positive, the second charge should be negative.
Thus, the value of the second charge is approximately − 4.18 × 1 0 − 12 C .
