Surface integral to Lateral integral?

[KingBigness]

Homework Statement

I have some working out my lecture gave me to a problem and I don't think I understand part of it. Hoping you could help me.


It's using Gauss' Law to find the capacitance of a cylindrical capacitor of length L but this information shouldn't matter for my question.

$\lambda=\frac{∂q}{∂l}$

$\oint E \bullet \Delta A = \frac{\sum q}{\epsilon}$

He then jumps to.

$E \int \Delta A = \frac{\lambda L}{\epsilon}$

First question, how does he go from the surface integral to the normal integral? Has it got anything to do with removing the dot product?

He then changes
$E \int \Delta A$ to E(2πrL)

I understand this is taking the integral of dA which becomes A, which is the area of a circle, hence 2πr, but how does he then bring the L into play...would this not be the Volume, not the area?

Thanks for any help.

 

[lanedance]

could it be that E is constant in the integral and always makes the same angle with the surface normal?

This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals

 

[KingBigness]

could it be that E is constant in the integral and always makes the same angle with the surface normal?

This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals

So when an angle doesn't change a surface integral can be turned into a lateral integral?
Sorry I haven't done much study on surface integrals so I don't know a lot about them. I'll look into them.

 

[lanedance]

so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A

 

[KingBigness]

so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A

Perfect sense, thanks so much for that