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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 12:48:17 IST
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A charge Q is placed
1. at the centre of the cube
2.at the centre of face
3.at the centre of edge
4.at one of its corner
find total electric flux through
a)The surfaces of cube
b)individual surfaces
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 12:56:23 IST
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gauss' law cant be implemented as charge lies on the surface
any other method i dont seem fruitful
rate me!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 12:59:19 IST
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q/epsilon knot for all the surfaces and q/6epsilon for individual surfaces
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:00:18 IST
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hey man its easy
center of cube q/epsilon0
center of edge q/4epsilon0
at one corner q/8epsilon0
hope i m right or i need a revision........
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:05:29 IST
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please explain
center of the edge=q/4epsilon 0
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:07:28 IST
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which gaussian surfaces r u using for determining the result????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:09:51 IST
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dont u remember that for determing E.F. thru a closed surface , no charge must lie on the gaussian surface
whats ur opinion
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:26:09 IST
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@nalgoel HEY DUDE , I THINK U CANNOT COMPLETELY VISUALISE THE QUESTION. SEE IF A CHARGE IS PLACED AT ONE FACE OF THE CUBE, IT IS IN CONTACT WITH ONLY THT FACE BUT NOT OTHER SURFACES. THEREFORE ALTHOUGH THE FLUX THROUGH THT FACE WILL BE ZERO BUT IT WILL PASS THROUGH OTHER 5 FACESAND SO V CAN VERY SAFELY APPLY GAUSS THEORUM. HOPE I HAD MADE MY POINT CLEAR. VOTE IF SATISFIED
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:29:23 IST
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@rahulraj
SEE, U WANT AN EXPLANATION FOR CHARGE KEPT AT CENTRE OF THE EDGE. NOW IF THE CHARGE IS KEPT AT THE CENTRE OF THE EDGE , ONLY Q/4 CHARGE WILL BE INSIDE THT CHARGE BECAUSE TO COMPLETELY COVER THT CHARGE V WILL REQUIRE FOUR SUCH CUBES.HENCE THE FLUX THROUGH ALL THE FACES WILL BE

THIS IS THE TOTAL FLUX PASSING THROUGH ALL THE FACES. NOW SINCE THE CHARGE IS IN CONTACT WITH TWO FACES , NO FLUX WILL PASS THROUGH THEM, HENCE THIS TOTAL FLUX WILL PASS THROUGH REMAINING FOUR FACES.
HENCE FLUX THROUGH EACH FACE WILL BE

hope u can understand the solution now
RATE IF SATISFIED OR U CAN ASK FURTHER.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 13:30:23 IST
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for charges at surface u can do like this assume two cubes and charge 2q at their common face. find total flux through both cube treating them as one object by gauss law and then divide the answer by 2 to get the flux by one cube(by symmetry).
similalry u can do for charge at the edge. at edge keep 8q charge coz a corner shared by 8 cube.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 14:12:19 IST
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gauss law can be easily apllied ...its impossible for human to solve this ques otherwise...for charges on edges and surface put four cubes and two cubes resp...just like we did it in solid state...charges on edges are shared by four equivalent cubes and on faces by two equivalent cubes...so gauss law and those structure in solid state are somewhat related...and similarly for corner wch is shared by 8 cubes and for centre its 1...rest can be done urslf...idea is this.....
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