How Does Gauss's Law Explain Electric Flux with External Charges?

[Noesis]

I have a question about electric flux.

Intuitively it makes that if a charge is outside a closed object, then the net flux inside that object will be zero..but once I write it down on paper it loses its sense.

I attached a picture for clarification.

Flux is equal to the field, times Area.

Flux = E*A

So, let's say we have a point charge on the side of a cylinder. For argument's sake, let's say on the axis of the cylinder.

For the flux to be zero, the flux entering on one side has to be equal to the flux exiting on the other side.

Now the problem with this in my mind, is that for the flux to be the same on both sides of the cylinder, both the field and the area must stay constant because of E*A.

Now, the area stays constant obviously...but the field drops off by an inverse square of the distance so it will have a lesser magnitude.

If the field changes...how in the world could the flux be zero?

I hope I explained my question decently...the picture should help.

Thank you anybody.

 

[Doc Al]

The net flux through all surfaces--the sides too!--must be zero. You can't just look at the flux through the two ends of the cylinder, you need to worry about the curved surface also.

 

[Noesis]

Hey thanks for responding Doc.

Ah...actually that might clear things up for me in a round about way.

I do know what you mean, that's why I chose the cylinder so the field could be parallel to the sides and there would be no flux...but that is me making the silly assumption that the field or "field lines" would continue to go in a parallel sense.

What if, as an idealization, there was an electric field with uniform direction, and it was perfectly perpendicular to the cylinder so the curved sides would experience no flux...would the flux still be zero then because of the discrepancy between the fields on both ends of the cylinder?

This might sound like a bit of a stretch...but there's actually a picture really similar to what I am asking in my book and it has made me think about it.

 

[Tomsk]

If you have a cylinder in a uniform electric field, with no flux through the curved sides, the flux is zero because you have to consider the edges of the cylinder as vectors. The direction of these vectors is along the normal to the area. The normal vectors would both be pointing outwards, in opposite directions, which is why they cancel and you get zero.

 

[Doc Al]

I do know what you mean, that's why I chose the cylinder so the field could be parallel to the sides and there would be no flux...but that is me making the silly assumption that the field or "field lines" would continue to go in a parallel sense.

Exactly.

What if, as an idealization, there was an electric field with uniform direction, and it was perfectly perpendicular to the cylinder so the curved sides would experience no flux...would the flux still be zero then because of the discrepancy between the fields on both ends of the cylinder?

If the field were uniform (thus the same at each end), the net flux would be zero. If the field were non-uniform (and thus different at each end, as in your example), the net flux would not be zero--implying that there is net charge within the cylinder.

This might sound like a bit of a stretch...but there's actually a picture really similar to what I am asking in my book and it has made me think about it.

If you can scan and post it, we can take a look.

 

[Noesis]

Thanks Doc Al...I appreciate the informative responses.

My questions have been answered and now I understand what my book meant by that awful picture.

Thanks again all of you guys.