Given uniform electric field E=5*10^3 i N/C,find the flux of this field through a square of 10 cm on a side whose plane is parallel to Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with X axis.

Question

Arun · Answer

Φ = E.ds
Given, side of the square = 10 cm = 10-1 m, E= 5* 103 i N/C
Hence, area of square = 10-2 i m2
Hence, Φ = 5 * 103 * 10-2 = 50 N m2 C-1
Now, Φ = E ds cosθ
Given, cosθ = cos 30 = 1.732/2 = 0.865
Hence, flux = 50 * 0.865 = 43.25 N m2 C-1

Khimraj · Answer

E = 5*10 3 N/C A = 0.01 m 2 ​flux = EAcos ​i) = 0 so flux = EA = 50 Nm 2 /C ​ii) = 30 so flux = EA /2 = 25 Nm 2 /C Hope it clears. If you like answer then please approve it.

Dipanshu Gautam · Answer

I am providing the concept of this numerical because above two solutions provided are totally wrong. In the second part of this numerical the angle b/w plane and X – axis is 30 degree . Since area vector is always perpendicular to plane . So angle b/w area vector and X- axis will be 60 degree. You can apply geometry by drawing the figure of this numerical.Hence, theta will surely be 60 degree.

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Given uniform electric field E=5*10^3 i N/C,find the flux of this field through a square of 10 cm on a side whose plane is parallel to Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with X axis.

Given uniform electric field E=5*10^3 i N/C,find the flux of this field through a square of 10 cm on a side whose plane is parallel to Y-Z plane. What would be the flux through the same square if the plane makes a 30° angle with X axis.

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