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Blazing goIITian

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4 Jun 2008 11:38:18 IST

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electric flux

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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?

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SUNDEEP ALLAMRAJU

Blazing goIITian

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4 Jun 2008 11:42:59 IST

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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.

ankit aggarwal

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4 Jun 2008 15:22:56 IST

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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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trueblue29

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7 Jun 2008 16:05:17 IST

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the flux though each face should be q/24enot cuz on completing the figure to get a guassian surface we get a large cube made of 8 cubes.
the flux thro' the sides containing the charge is zero.
the large cube has 6 faces..each made of 4 sides of the smaller cubes..therefore flux thro each small side is eaual to q/(4*6*enot)=q/24enot.

C!-!!NMAY

Blazing goIITian

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7 Jun 2008 18:36:15 IST

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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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akshay A NEW BEGINNING...

Blazing goIITian

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7 Jun 2008 20:30:33 IST

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it is q / 24e not

Gowtham S

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7 Jun 2008 20:32:56 IST

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q/24e e=epsilon

Ebrahim F. Attarwala

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7 Jun 2008 21:14:56 IST

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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

Rate me if you agreed with my answer.

Ashwin Menon

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9 Jun 2008 01:21:36 IST

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can anyone provide a diagram for i did not understand

VIV PAR

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9 Jun 2008 13:56:19 IST

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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.

Ashwin Menon

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10 Jun 2008 19:59:30 IST

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thanks!

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