How Does an External Charge Affect Flux Through a Cube's Face?

[jd12345]

A charge q is kept at the centre O of cube of length L(ABCDEF). Another charge q is kept at a distacne L from the centre O. Find the flux through face ABCD

Well the flux through whole of the cube will remain unchanged that is q/ε.
But now since we have another charge q outisde the cube we cannot use symmetry and say flux through each face is q/6ε.
I have no idea how to proceed ahead. please help. thank you

since i don't know how to post the image ill just give a description about it :- charge q which is outside the cube is kept such that the face of the cube bisects the distance between the two charges

 

[tiny-tim]

hi jd12345!

charge q which is outside the cube is kept such that the face of the cube bisects the distance between the two charges

just as if another cube was there!

 

[jd12345]

lol yeah - it could be written in that way too
I always make things more complicated - damn it

 

[Hassan2]

lol yeah - it could be written in that way too
I always make things more complicated - damn it

Is the cube a conductor? If yes, is it grounded?

 

[Philip Wood]

Isn't there more to tiny-tim's post than an alternative description of the set-up? Isn't he hinting at how the problem is solved – in a single line? [Think about what the component of E perpendicular to the face at each point will be, by adding the fields due to the individual charges at that point.]

I first misread the question as the charges being equal and opposite. The result is just as easy to find, and at least as interesting.

 

[Hassan2]

The result is just as easy to find, and at least as interesting.

I haven't figured it out yet. The distribution of the flux density inside the cube is still symmetric due to the box ( which is equi-potential). So the flux due to the inner charge is still Q/6ε. However, determining the flux due to the charge outside the cube doesn't seem to be trivial even if there was no charge in the cube. Any hints please?

 

[jd12345]

So the flux due to the inner charge is still Q/6ε.

i don't think flux due to inner charge will be Q/6ε. The charge outisde will alter the electric field lines of the inner charge and hence electric field on all the faces will not be the same - so flux not same on all faces

But i still cannot solve the problem. Do i imagine another cube as tiny-tim hinted? How will that help?

 

[Bipolarity]

IF the charges are the same sign, shouldn't the flux through the face be zero due to reflective symmetry about the cube face ABCD?

 

[Hassan2]

i don't think flux due to inner charge will be Q/6ε.

In a simple language, the box shield the inner charge so that it doesn't feel its existence of the outer charge. Hence the field inside the box is distributed as if there was no charge outside.

Mathematically speaking, the solution of Laplace's equation for potential with a charge in the center and a constant boundary condition on the cube is symmetric.

 

[jd12345]

Sorry didnt tell you before but the answer is given at the back of my text - its zero
any thoughts on why its zero?

 

[Hassan2]

IF the charges are the same sign, shouldn't the flux through the face be zero due to reflective symmetry about the cube face ABCD?

If both charges were in identical cubes, the symmetry would be obvious but the problem doesn't have such a symmetry.

In my opinion, some negative charge of opposite sign of q is accumulated on face ABCD. This makes the flux non-zero.

 

[Hassan2]

Sorry didnt tell you before but the answer is given at the back of my text - its zero
any thoughts on why its zero?

Perhaps the cube is not a conductor?!

 

[tiny-tim]

hi jd12345!

(just got up :zzz:)

Sorry didnt tell you before but the answer is given at the back of my text - its zero
any thoughts on why its zero?

perhaps you're misreading the queston?

the way i read it, it's only asking for the flux through one face, the common face between the two cubes

from symmetry, one flux will obviously be minus the other! ​

The charge outisde will alter the electric field lines of the inner charge

yeees, but it's a lot easier to think of the fields as being completely separate

electromagnetic fields don't interact with each other, they just add

(btw, "shielding" is irrelevant)

so you can always calculate each electric field (or flux) separately, and add them

 

[jd12345]

ok thanx

 

[Philip Wood]

And for my misreading (thinking the charges were equal and opposite) the flux through the common face is q/3ε₀.

 

[Hassan2]

yeees, but it's a lot easier to think of the fields as being completely separate

electromagnetic fields don't interact with each other, they just add

(btw, "shielding" is irrelevant)

so you can always calculate each electric field (or flux) separately, and add them

Thanks. This makes things easier.