Consider a uniform electric field E = 3 × 103 îN/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
(a) Electric field intensity, = 3 × 103 î N/C
Magnitude of electric field intensity, = 3 × 103 N/C
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s2 = 0.01 m2
The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0°
Flux (Φ) through the plane is given by the relation,
Φ =
= 3 × 103 × 0.01 × cos0°
= 30 N m2/C
(b) Plane makes an angle of 60° with the x-axis. Hence, θ = 60°
Flux, Φ =
= 3 × 103 × 0.01 × cos60°
= 15 N m2/C
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Right ans thanks
Great answer...thank u and it's easy to understand
How it came 60°.since it is a square it should be 90°is it.then the answer is wrong right
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