Question 15

Consider a uniform electric field E = 3 × 10^{3} 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?

Answer

(a) Electric field intensity, = 3 × 10^{3} î N/C

Magnitude of electric field intensity, = 3 × 10^{3} N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = s^{2} = 0.01 m^{2}

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 × 10^{3} × 0.01 × cos0°

= 30 N m^{2}/C

(b) Plane makes an angle of 60° with the x-axis. Hence, θ = 60°

Flux, Φ =

= 3 × 10^{3} × 0.01 × cos60°

= 15 N m^{2}/C

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Md Danish
2019-06-27 21:31:31

Right ans thanks

rigio taditya
2018-07-17 18:29:56

Great answer...thank u and it's easy to understand

Neshvin
2018-03-17 10:11:56

How it came 60Â°.since it is a square it should be 90Â°is it.then the answer is wrong right

Anu
2018-03-02 23:30:10

Useful!

Deeksha
2016-04-26 17:01:38

Thank u soo much for this wonderful website!

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