Assume that at some point just outside the surface of the conductor, the electric field has magnitude E and is directed toward the surface of the conductor. What is the charge density η on the surface of the conductor at that point?

η = -E⋅ε0

Explanation:

(1) Use Gauss's law with a short and flat cylindrical surface

(2) Calculate the flux through the top of the cylinder

Express your answer in terms of A, E, and any needed constants.

→ φtop = -EA

(3) Calculate the flux through the bottom of the box

Answer in terms of A, E, and any needed constants.

→ Φbot = 0

(4) Find the net charge qin inside this Gaussian surface.

→ qin = ηA

(5) Now apply Gauss's law, neglecting any contribution to the flux due to the very short sides of the cylinder. Gauss's law states that ϵ0⋅ΦE=qin. The area A should cancel out of your result.

→ η = -E⋅ε0

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false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.