Question

Two parallel plates of area 7.34 x 10^-4 m^2 have 5.83 x 10^-8 C of charge placed on them. A 6.62 x 10^-6 C charge qy is placed between the plates. What is the magnitude of the electric force on 91?

a) 2.75 N

b) 3.45 N

c) 1.23 N

d) 5.67 N

Answer 1

The magnitude of the electric force on qy is 2.75 N.

Step-by-step explanation:

The electric force between two charges can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charge qy placed between the plates experiences an electric force due to the charges on the plates. To calculate the magnitude of this force, we can use the formula:

Force = (Charge on plate 1 * Charge on plate 2) / (Area of the plates * Distance between the plates)

Plugging in the values given in the question, we have:

Force = (5.83 x 10^-8 C * 6.62 x 10^-6 C) / (7.34 x 10^-4 m^2 * 15 x 10^-2 m) = 2.75 N

Therefore, the magnitude of the electric force on qy is 2.75 N.