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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.
one year ago

Arun
24490 Points
							Φ = E.dsGiven, side of the square = 10 cm = 10-1 m, E= 5* 103 i N/CHence, area of square = 10-2 i m2Hence, ​Φ = 5 * 103 * 10-2 = 50 N m2 C-1Now, ​Φ = E ds cos​θGiven, cos​θ = cos 30 = 1.732/2 = 0.865Hence, flux = 50 * 0.865 = 43.25 N m2 C-1

one year ago
Khimraj
3008 Points
							E = 5*103 N/CA = 0.01 m2​flux = EAcos$\Theta$​i) $\Theta$ = 0so flux = EA = 50 Nm2/C​ii) $\Theta$ = 30so flux = EA$\sqrt{3}$/2 = 25$\sqrt{3}$ Nm2/C Hope it clears. If you like answer then please approve it.

one year ago
Dipanshu Gautam
24 Points
							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.

3 months ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions