
#1
Feb2013, 12:05 AM

P: 78

In calculating the electric field bw two parallel plate i.e a capacitor I encountered a problem :where to take the Guassian surface..When I consider the Guassian surface.a cube bw the two plates Electric intensity becomes 0..But when I take the Guassian surface with a portion of capictor inside the surface I get the answer..But it is confusing..How may I know where to take the Guassian Surface?




#2
Feb2013, 01:41 AM

P: 2,074

You can choose whatever surface you want. Of course, you probably would want to choose a surface that makes Gauss's law actually useful. For example, picking a surface which is in between the two plates, the total integrated flux around the surface must be 0 (no enclosed charge) and this is true because whatever flux enters one side of the box, leaves the box on the other side.
Picking this surface in the center, you get an equation 0=0, which is of no help to you, but is a true equation nontheless. 



#3
Feb2413, 01:18 AM

P: 78





#4
Feb2413, 01:45 AM

P: 7

Guassian Law:A confusion
a surface to be chosen as Gaussian surface where all of the electric field lines are perpendicular to that surface.....
remember.. flux=ExA cos(angle bw area vector of the surface and electric field lines) if u like the answer then support 



#5
Feb2413, 02:44 AM

P: 252

When calculating flux you also have to pay attention to the signs. The flux means the net flux or the net outflow, if you will. If as much comes in as goes out, the total flux is zero. If you take a cube between two charged plates, there is the same electric field coming in from the bottom as is going out the top, so the net outflow is zero. Gauss' law doesn't give the electric field inside it, only the total flux out of it, but which can sometimes be used to get the actual electric field.




#6
Feb2413, 02:57 AM

P: 8

then,why is field lines always perpendicular to the surface we choose?




#7
Feb2413, 01:12 PM

P: 197

The first step in choosing Gaussian surface is learning Gauss's Law. It says that the electric flux through any closed surface is proportional with the charge enclosed. Then you won't be perplexed by the fact that flux is zero anywhere between the plates, because there are no any charges there.



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