Question

The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of 1.5 m2, while surface 2 has an area of 3.4 m2. The electric field E in the drawing is uniform and has a magnitude of 190 N/C. It is directed towards the two perpendicular surfaces, making an angle 35o with the bottom surface. Find the electric flux through (a) surface 1 and (b) surface 2.

Answer 1

Remember electric flux is given by:

= E.A

= E*A*cos theta

Where E = electric field = 190 N/C

A = area of surface

theta = angle between electric field and surface in the direction of electric field (Important)

Part A.

For surface 1

A1 = Area of surface 1 = 1.5 m^2

theta1 = 35 deg

So,

1 = E*A1*cos theta1

1 = 190*1.5*cos 35 deg

1 = 233.5 N-m^2/C

Part B.

For surface 2

A2 = Area of surface 2 = 3.4 m^2

theta2 = 90 - 35 = 55 deg (Since surface 1 and surface 2 are perpendicular)

So,

2 = E*A2*cos theta2

2 = 190*3.4*cos 55 deg

2 = 370.5 N-m^2/C

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