Question

Two parallel plates of area 7.34x10-4 m^2 have 5.83x10-8 C of charge placed on them. A 6.62x10-6 C charge q1 is placed between the plates. What is the magnitude of the electric force on q1?

a. 2.01 × 10^-1 N
b. 5.42 × 10^-4 N
c. 8.73 × 10^-3 N
d. 3.28 × 10^-2 N

Answer 1

To find the electric force on the charge q1 between parallel plates, calculate the surface charge density (sigma = Q / A), find the electric field (E = sigma / epsilon_0), and then use the electric field to determine the force (F = q1 * E). Perform these calculations to determine the correct answer.

Step-by-step explanation:

To determine the magnitude of the electric force on the charge q1 placed between two parallel plates, we must first find the electric field E between the plates. The electric field E can be calculated from the surface charge density σ, which is the charge per unit area on the plates (σ = Q / A), where Q is the total charge on one plate and A is the area of the plate.

Then, using the formula E = σ / ε0, where ε0 is the vacuum permittivity (ε0 = 8.85 x 10-12 C2/N·m2), we can find E. Once E is known, the force F on q1 can be found using F = q1E.

Let's calculate the surface charge density first:

σ = (5.83 x 10-8 C) / (7.34 x 10-4 m2)

Now, we calculate the electric field E using σ and ε0:

E = σ / ε0

Finally, we use the electric field to find the force on q1:

F = (6.62 x 10-6 C)(E)

Without the numerical calculations complete, we don't have the exact answer, but the methodology is outlined here for you to follow and find the correct magnitude of the electric force on q1.