IIT JEE , AIEEE Forum - Physics , Electricity

neeraj_agarwal_1990

consider a conducting box with an elliptical cavity in it...
a charge q is placed at the center of the cavity....

now we have to find the flux through the surface of the cavity.....

according to me it should be 0 because a charge -q is induced at the inside surface and the enclosed charge becomes 0...

but answer given is q/ep. and i have checked at 2-3 places where answer is q/ep.
so answer doesn't seem to be wrong...but where am i going wrong??

ashgirl

yes it is q/E_0.

since due to induction, in the inner part of the cavity -q charge is induce and on the outer part of conducting sphere q is induced.

the flux coming out of box is q/E₀due to q.

harsh15

The answer is q/ep.

The probable reason is that under electrostatic condition no excess charge can exist inside the body of the conductor.So the charge given comes on the surface.

Now consider a gaussian surface enclosing the conductor.the total charge inside the gaussian surface is q.

hence the flux through the surface is q/ep.(by guass's theorem).

ashgirl

see, E₀ is epsilon !! and not electric field!!

harsha_27

you should not consider the induced charge while using gauss's law ............
because that induced charge has no independent existence .......i mean it is not real charge.....it came into existence by the charge q itself, when u have considered the original charge q .....u need not (indeed u should not)again consider the induced charge produced by it......got it?

pardesi

well there is no problem in considering induced charge as such because every positive will have a negative charge.Gauss' law talks about the net charge enclosed in a surface no matter how the charge originated

harsha_27

While applying Gauss's law we simply shouldn't add up whatever charges we see in the area.......the charges must be independent in existence ......if u doesn't agree with me , tell me if we remove that q charge out of the cavity do u say the net flux enclosed by the cavity now is -q/

0???

pardesi

well nowhere in the proof of gauss's law is it written that charges should have "independent existance".
if u remove that charge q then the -q will also 'disappear' due to rearrangement

harsha_27

flux implies only net amount of electric field lines enclosed by the region ......here -q charge induced by the q charge doesn't produce electric field lines independently......that is it is not real charge.....

pardesi

what is 'real' or 'imaginary' charge
any charge has field assicciated with it and in partciular this charge cancels out any net field inside the conductor

harsha_27

well i've not mentioned imaginary charge anywhere in my msg..................
by real charge i mean the charge which has an independent existence and can produce an electric filed lines of its own ........induced charge is not an independent charge....why don't u understand?

pardesi

".i mean it is not real charge.....it came into existence by the charge q itself, when u have considered the original charge q .....u need not (indeed u should not)again consider the induced charge produced by it......got it?"

i was taught in my school not real means imaginary so...
Who said NOT REAL charges don't have field lines.
they have and please go through the proof of gauss's law and wherever u studied about these NOT REAL/NOT INDEPENDNT charges.
charges are bound to produce fields that's how there is 0 field inside a conductor.
If we ever go by what u said then no matter what happens if there is REAL charge in side a conductor then no matter how it rearranges ther will be a net field

harsha_27

u r totally confused b/w net flux and net field........thats it....i'm unable to make it clear to u online

pardesi

it's good that u r online otherwise i don't knoe what would have happened to my concepts.
consider a metallic sphere having a cavity inside .Let the cavity have a charge q hanging inside let's take a
concentric sphere inside the sphere as gaussian surface clearly field every where on the sphere is 0 so flux is 0
But according to u since nly REAL charges have field lines so the flux will be non zero

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