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19 Jun 2007 20:41:43 IST
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Find the flux through any one face of a cube of side "a" when a charge q is placed at any one corner of the cube.
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There is no better feeling in this world than being a winner! |
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19 Jun 2007 22:02:25 IST
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it is 1/6(q/4  ) = q/24 it will require 3 more cubes to cover whole cube. so flux linked with whole cube is q/6 . &flux linked with each face is 1/6 of it i.e. q/24 . |
Himanshu Jain
P.E.C.
http://jainhim.blogspot.com/ |
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19 Jun 2007 22:50:06 IST
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Here's the solution - u must hav seen the que. , where charge q is placed at the centre of the cube , and flux through one face is q/6E (E is epsilon ) . Now , visualize that the point charge is placed at the centre of a large cube made of 8 small cubes . Now , accordingly , flux through open face is q/6E . But we require flux through one fourth of this large side (which is one side of the given small cube ). That will be 1/4(q/6E) , i.e. , q/24E I hope u got it !!!!!!!
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Umang |
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20 Jun 2007 12:25:58 IST
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I agree wid them...
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20 Jun 2007 13:28:43 IST
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See, if we place eight cubes more the flux=q/epsilon0
.: through one cube=flux/8
.: through one face should be flux/48????
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There is no better feeling in this world than being a winner! |
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20 Jun 2007 21:59:07 IST
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Hey joyfrancis ! I am not getting what u r trying to say . Hav u understood my solution ?????
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Umang |
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1 Aug 2007 21:59:07 IST
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1 / 6 of q/4 Eo = q/24 Eo
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- Gaurav Ragtah (spideyunlimited)
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1 Aug 2007 23:14:57 IST
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him26.89 or watever ur name is, It wud not require 3 but 7 more cubes to cover up the charge. Think over it... @umang I dont understand ur logic of 1/4th of this large side??? @joyfrancis Though there will be 8 cubes, only three faces for each cube will be responsible for the flux as their area vector will be perpendicular to the electric field. Hence the answer will be twice of wat u said And hence q/24E is the correct answer as umang and him said.but their logic comes 4m nowhere...
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HOPE U GOT IT... |
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