A charge Q is placed at the corner of a cube of edge a . what is the electric flux thru each of the cube faces?

The flux thro', the faces containing the corner at which charge is placed is 0.The flux thro' the other three faces is q/24e₀ as the net contribution should be q/8e₀.

as the charge is placed at the corner net flux through the cube is   q/e₀

since there are 8 faces

therefore flux through each face should be  q/8e₀

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see as the charge is kept at the corner of cube of edge a,so flux passing through the face containing corner is zero,,,,,but the flux passing through each others faces is 0 and ,,,net flux is 0 i.e. 0..

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Flux on the opp face is Q/24 epsilon.

consider 8 cubes. so flux at one of its face will be Q/6 Epsilon. and one face is consist of 4 small cubes. so net flux on the required face will be Q / 24 Epsilon.

Flux on the face in which it is placed will be Zero. because electric lines of force will be in the direction perpendicular to the area vector which is normal to the surface area. so net flux will be equal to 0

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CONSIDER THE CUBE AS SHOWN AND A CHARGE AT ITS CORNER..TO COMPLETE THE GAUSSIAN SURFACE YOU NEED SEVEN MORE CUBES.THUS THE NET FLUX THROUGH WACH CUBE WILL BE Q/8.

(NOW IMAGINE BIG CUBICAL ROOM AT THE CORNER OF WHICH IS PLACED A LIGHT SOURCE(IN OUR CASE ITS Q/8)).CLEARLY THE LIGHT CANT PASS PARALLEL WAYS SO IT WILL PASS THROUGH THE THREE OPPOSITE FACES.HENCE THE TOTAL FLUX IN OUR CASE WILL BE DIVIDED BY THREE.IE.Q/24.