A point charge Q lies at the center of a hollow cube of side L. What is the tota

Question

A point charge Q lies at the center of a hollow cube of side L. What is the total electric flux through any single face of the cube? Returning to the point-charge configuration of Problem (1), compute the total outward flux (of the electric field) through a spherical surface centered at the origin, with radius R > L. An infinite, flat sheet of charge has uniform surface-charge density sigma. A disk of radius R is placed at a distance D from the sheet. The surface normal to the disk makes an angle theta with the surface normal to the sheet. What is the magnitude of the electric flux that penetrates the disk?

Explanation / Answer

Answer:

(a) According to Gauss law of electrostatics, electric flux through a close surface is Q/0 .

And the surface area of the cube is 6 times the area of one surface.

Electric flux lines are spreading symmetrically through the all six faces of the cube, Therefore, electric flux E passing through any one face is one sixth of the total flux through closed surface.

Therefore, elctric flux passing through any one surface flux through one surface=1/6(Q/0) =Q/ 60 .

(b) You need to give the point-charge configuration in problem(1), then only we can slove this problem.

(c) Electric fiel due to an infinite, flat sheet of charge has uniform surface-charge density is /20 .

Given that, a disk of radius R is place at a distance D from the sheet and the angle between the two surface nomals is .

Therefore, the magnitude of the electric flux that penetrates the disk is E = EA cos.

Here A is the area of the gievn disk, area A = R2 .

Hence, E = EA cos = (/20) (R2) cos.

If = 00, the eletric flux is maximum, that means E = EA = (/20) (R2)    (since cos 00 = 1)

If = 900, E = EA = 0 . Since Cos 900 is zero.

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