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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.

Grade:12

3 Answers

Arun
25763 Points
3 years ago
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 cos​θ
Given, cos​θ = cos 30 = 1.732/2 = 0.865
Hence, flux = 50 * 0.865 = 43.25 N m2 C-1
Khimraj
3007 Points
3 years ago
E = 5*103 N/C
A = 0.01 m2
​flux = EAcos\Theta
​i) \Theta = 0
so flux = EA = 50 Nm2/C
​ii) \Theta = 30
so flux = EA\sqrt{3}/2 = 25\sqrt{3} Nm2/C 
Hope it clears. If you like answer then please approve it.
Dipanshu Gautam
24 Points
one year ago
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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