Question

A hollow tubular box resting within a horizontal electric field of magnitude E=8,954 N/C as shown in the figure, has bases with total radius of 30cm, while the radius of its inner hollow circle is 18cm
Calculate:
a) The electric flux through surface A (notice surface A is donut shaped).

b) If surface B is the opposite side of the box (which is also donut shaped) what should be the electric flux on it? Explain WHY.

Answer 1

A. Here the electric field is normal to the surface A and B of the box. Area of the surface d =지성-r) rOuter radius Area of the surface π( -r =Inner radius] 3.14x(0.30 2) = -0.18 = 0.181 m. = 8954 x 0.181 1619.45 Nm 2/C B. Electric flux on the surface B is negative that at A because at A electric field and area vectors are antiparallel to each other. It follows from the convention that outward normal is taken as positive direction of area vector