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

change tension

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this