Question

Find the electric flux (in N m2/c) through a rectangular area 5.0 cm x 4.0 cm between two charged parallel plates where there is a constant electric field of 27 N/C for the following orientations of the area. (Enter the magnitudes.) (a) parallel to the plates N m2/c N m2/c N m2/c (b) perpendicular to the plates (c) the normal to the area making a 30 angle with the direction of the electric field; note that this angle can also be given as 180 30

Answer 1

Answer:

a) 0.054 N m²/C

b) 0

c) 0.0467 N m²/C

Explanation:

a) Electric flux = Φ = E A cos θ

E = Electric field = 27 N/C

Area = A = 0.05×0.04 =0.002 m²

Angle = θ = 0 , since they are parallel.

Φ= Electric flux = (27)(0.002)(cos 0) = 0.054 N m²/C

b) When θ= 90°, perpendicular

Ф = Electric flux = (27)(0.002)(cos 90) =0

c) θ = 30 degrees,

Φ = Electric flux = (27)(0.002)(cos 30) = 0.0467 N m²/C

Answer 2

The electric flux through the plates at parallel orientation is 0.054 Nm²/C.

The electric flux through the plates at perpendicular orientation is 0.

The electric flux through the plates at 30 degrees orientation is 0.047 Nm²/C.

Area of the coil

The area of the coil is calculated as follows;

A = 0.05 x 0.04 = 0.002 m²

Electric flux

The electric flux through the plates is calculated as follows;

Ф  = EA cosθ

When the field is parallel to the plate

Ф  = 27 x 0.002 x cos(0)

Ф = 0.054 Nm²/C

When the field is perpendicular to the plate

Ф = 27 x 0.002 x cos(90)

Ф  = 0

When the field is 30 degrees to the area of plate

Ф = 27 x 0.002 x cos(30)

Ф = 0.047 Nm²/C

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