The length of each side of a cubical closed surface is l. If charge q is situated on one of the vertices of the cube,then find flux passing through shaded face of cube.
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let the cube be ABCDEFGH with ABCD being one face and EFGH being the symmetrically opposite face. Let us find the electric flux thru face EFGH.
Due to the charges at E,F, G or H the flux thru the face EFGH = 0.
Due to charge q at A, the flux that flows through inside the cube is 1/8th of total flux due to q at A. So that is = 1/8 * (q/ε) using the Gauss law. This flux goes into three faces that dont contain A ie., BCFG, EFGH, DHCF. So the flux due to charge q at A passing thru face EFGH = 1/24 * (q/ε).
Total flux due to charges at A, B, C and D passing thru face EFGH
= 1/6 * (q/ε)
Due to the charges at E,F, G or H the flux thru the face EFGH = 0.
Due to charge q at A, the flux that flows through inside the cube is 1/8th of total flux due to q at A. So that is = 1/8 * (q/ε) using the Gauss law. This flux goes into three faces that dont contain A ie., BCFG, EFGH, DHCF. So the flux due to charge q at A passing thru face EFGH = 1/24 * (q/ε).
Total flux due to charges at A, B, C and D passing thru face EFGH
= 1/6 * (q/ε)
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