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A uniform electric fieldˆˆ ˆ 4E i j k= + + exists in space. Find out the electric flux through asquare ring of edge length 3 m placed in x-z plane

Question

A uniform electric fieldˆˆ ˆ 4E i j k= + + exists in space. Find out the electric flux through asquare ring of edge length 3 m placed in x-z plane

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Solution

The electric flux through a surface is given by the dot product of the electric field and the area vector. The area vector is perpendicular to the surface and its magnitude is equal to the area of the surface.

Given that the electric field is E = 4i + 4j + 4k and the square is in the x-z plane, the area vector of the square will be along the y-axis (since it is perpendicular to the x-z plane).

The area of the square, A = side^2 = 3m * 3m = 9 m^2. So, the area vector A = 9j.

The electric flux φ is given by φ = E . A = (4i + 4j + 4k) . (9j) = 4*9 = 36 N.m^2/C.

So, the electric flux through the square ring is 36 N.m^2/C.

This problem has been solved

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