Question

A test charge of +2 μC is placed halfway between a charge of 5.5 μC and another of 4.5 μC, separated by 10 cm. What is the magnitude of the force on the test charge in Newtons?

a) 1.35 N
b) 2.25 N
c) 3.15 N
d) 4.50 N

Answer 1

The magnitude of the force on the test charge is approximately 0.1 N.

Step-by-step explanation:

The magnitude of the force on the test charge can be calculated using Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the test charge is halfway between a charge of +5.5 μC and another of +4.5 μC, separated by 10 cm. We can calculate the force using the equation:

F = k * (q1 * q2) / d^2

where F is the force, q1 and q2 are the charges, d is the distance, and k is Coulomb's constant, which is approximately 9 x 10^9 Nm^2/C^2.

Plugging in the values, we have:

F = (9 x 10^9) * ((2 x 10^-6) * (5.5 x 10^-6)) / (0.1^2)

F = 0.099 N, which is approximately equal to 0.1 N. Therefore, the magnitude of the force on the test charge is approximately 0.1 N.