Question

A uniform electric field with a magnitude of 6 × 10^6 N/C is applied to a cube of edge length 0.1 m as seen in Fig 22-2 above. If the direction of the E - field is along the +x-axis, what is the electric flux passing through the shaded face of the cube?

Answer 1

In fig- n( cap) = area vector , which is perpendicular to the surface and directed radially outward.

since electric field is along + x- axis ( as shown in fig) , so electric flux will be only for the left and right face of the cube. for other faces ( top, bottom, front, back) theta = 90 so flux = 0.

since fig is not given in the question , if shaded area along + x- axis ( ie along right face) as shown in fig:

flux = E. ds

       = E *ds * cos 0                              ( theta= 0 ,because E and area vector are in same direction)

       = 6*10⁶ * ( 0.1)² * 1

       = 60*10³ N m² C^-1

or

if shaded area is toward left face, then

flux = E*ds*cos 180

      = - 60*10³ N m² C^-1