Question

The surface charge density on an infinite charged plane is -1.60 × 10⁻⁶ C/m². A proton is shot straight away from the plane at 1.50 × 10⁶ m/s. What is the force experienced by the proton due to the electric field created by the charged plane?

Answer 1

The force experienced by the proton due to the electric field created by the charged plane is -2.89 x 10^-17 N.

Step-by-step explanation:

To calculate the force experienced by the proton due to the electric field created by the charged plane, we can use the following formula:

F = q * E

Where F is the force, q is the charge of the proton, and E is the electric field. The charge of the proton is +1.602 x 10^-19 C and the electric field can be calculated using the surface charge density:

E = sigma / epsilon_0

Where sigma is the surface charge density (-1.60 x 10^-6 C/m^2) and epsilon_0 is the permittivity of free space (8.85 x 10^-12 C^2/N*m^2).

Substituting the values, we have:

E = (-1.60 x 10^-6 C/m^2) / (8.85 x 10^-12 C^2/N*m^2)

E = -180.79 N/C

Now, we can calculate the force:

F = (1.602 x 10^-19 C) * (-180.79 N/C)

F = -2.89 x 10^-17 N

Therefore, the force experienced by the proton is -2.89 x 10^-17 N.