Gauss's Law and Finding the electric field

AI Thread Summary
The discussion focuses on applying Gauss's Law to find the electric field around two long charged concentric cylinders. The charge per unit length is provided for both cylinders, with the inner cylinder positively charged and the outer cylinder negatively charged. The key formula used is EA = qenc/ε0, where E is the electric field, A is the surface area, and qenc is the enclosed charge. The confusion regarding the height of the cylinder is clarified, noting that it cancels out when using the surface area of the cylinder in the calculations. The participants confirm that understanding the correct area is crucial for solving the problem accurately.
musicmar
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Two long charged, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 5.0 x 10-6 C/m on the inner cylinder and -7.0 x 10-6 on the outer cylinder. Find the electric field at (a) r = 3.0 cm and (b) r = 8.0 cm, where r is the radial distance from the common central axis.

Homework Equations


ε0 ∮EdA = qenc
λ= Q/L


The Attempt at a Solution



EA = qenc/ε0


E = λL/ π r2 ε0

From here, I don't know what to do with L. The height of the cylinder should be irrelevant, right?
I have another similar question using ρ and volume, leaving me with a height in the formula for the volume that I don't know what to do with.

Thanks.
 
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EA = qenc/ε0

this is the right formula, just make it clear, what A is.

ehild
 
I figured out what I was doing wrong. The area is the surface area of the cylinder, not the area of the base. So, L cancels when (2 pi r)h is in the denominator.
 
musicmar said:
I figured out what I was doing wrong. The area is the surface area of the cylinder, not the area of the base. So, L cancels when (2 pi r)h is in the denominator.

Well done!

ehild
 
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