In the Week 8 tutorial we used a Gaussian cylinder to determine the electric fie

Question

In the Week 8 tutorial we used a Gaussian cylinder to determine the electric field of an infinite sheet of charge to be... Show that a Gaussian cube (with sides of length = s) would have the same result. What features make the cylinder and cube good choices for this problem? Explain why a Gaussian sphere is not a suitable choice for this problem. A cone with a height of 4 m and a radius of 2m is placed in an electric field E and is oriented such that base is perpendicular to the field. Determine the electric flux in terms of E through the slanted portion of the cone. (The lateral face ... all but the base.)

Explanation / Answer

a) Using Gauss Law,

E . A = Qin / e0

Qin = (sigma * s^2)

and A of surface = s^2

E ( s^2) + E(s^2) = (sigma s^2) / e0

E = sigma s^2 / (e0 2s^2) = sigma / 2e0

b) field parallel to sheet will be zero.

so flux through the surfaces that are perndicular to the sheet will have
zero flux through them.

(four faces of cube and curved surface of cylinder)

and two surface are paralllel to sheet, and every point on this surface is equidistant from sheet.
so field is same throughout these surfaces.

hence total flux will be 2E.A

c) but in case of sphere, fields will be different for every point (as they are not equidistant from sheet)

hence field cannot be find from flux as field is not constant

2. projected are = pi r^2

flux = E.A = E ( pi r^2) = 12.57E or 4 pi E

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