Question

Two point charges, one with a magnitude of -2.0 μC and the other with a magnitude of 4.0 μC, are separated by a distance L and experience an electric force of 11.25 N. What is the value of L, measured in meters?

A) 0.100 m
B) 0.200 m
C) 0.300 m
D) 0.400 m

Answer 1

Using Coulomb's Law, the separation distance L between the two point charges is calculated to be approximately 0.200 m. This is option B) on the provided list.

Step-by-step explanation:

The question requires the application of Coulomb's Law to find the separation distance L between two point charges experiencing an electric force. According to Coulomb's Law:

F = k * |q1 * q2| / L^2

where:

F is the force between the charges,

k is Coulomb's constant (≈ 8.99 x 10^9 N m^2/C^2),

q1 and q2 are the magnitudes of the charges, and

L is the distance between the charges.

Substituting the known values into this equation:

11.25 N = (8.99 x 10^9 N m^2/C^2) * (2.0 x 10^-6 C) * (4.0 x 10^-6 C) / L^2

We can solve for L, squaring 11.25 N and dividing by the product of the charges and Coulomb's constant, then taking the square root:

L = sqrt((8.99 x 10^9 N m^2/C^2 * 2.0 x 10^-6 C * 4.0 x 10^-6 C) / 11.25 N)

After calculating, this yields:

L ≈ 0.200 m, which matches option B) 0.200 m.