Question

The figure below shows a closed cylinder with cross-sectional area A = 2.60 m².

The upper and lower circular surfaces of a vertically-oriented cylinder are labeled A. Electric field vector E points vertically upward, going up through the lower surface, up through the cylinder, and finally pointing up through the upper surface.

The constant electric field E has magnitude 1.85 ✕ 10³ N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux (in (N · m²)/C) through the cylinder's top and bottom surfaces.

(a) top surface

  

N · m2

C

(b) bottom surface

  

N · m2

C

(c) Determine the amount of charge (in C) inside the cylinder.

C

Answer 1

Elwctric flux is calculated by following formula :

where, E = magnitude of electric field ; S = surface area ; = angle between electric field and area vector

Given,

E = 1.85 x 10³ N/C directed towards top surface

Area of bottom and top surface of cylinder = 2.60 m²

On top surface, ,

So,

On bottom surface, ,

So,

On curved surface, ,

So,                   

The net electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.