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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

high-sugar diet and high cholesterol