How Is Electric Flux Calculated Through a Cylindrical Surface?

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SUMMARY

The electric flux through a cylindrical surface defined by \(x^2 + y^2 = R^2\) for \(0 \leq z \leq h\) is calculated using the electric field \(\vec{E} = 5 \lambda \frac{x \vec{i} + y \vec{j}}{x^2 + y^2}\) resulting from a linear charge density \(\lambda\). The flux through the top and bottom surfaces of the cylinder is zero due to the absence of the \(k\) component in \(\vec{E}\). The relevant integral for the cylindrical surface normal must be computed, and it is recommended to convert to cylindrical coordinates for simplification.

PREREQUISITES
  • Understanding of electric flux and its mathematical representation.
  • Familiarity with vector calculus, specifically dot products and surface integrals.
  • Knowledge of Gauss's theorem (divergence theorem) for volume integrals.
  • Proficiency in converting Cartesian coordinates to cylindrical coordinates.
NEXT STEPS
  • Study the application of Gauss's theorem in electrostatics.
  • Learn how to compute surface integrals in cylindrical coordinates.
  • Explore the concept of electric flux in different geometrical configurations.
  • Review vector calculus topics related to divergence and curl.
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Students and professionals in physics and engineering, particularly those focusing on electromagnetism and electrostatics, will benefit from this discussion.

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Homework Statement



The z-axis carries a constant electric charge density of λ units of charge per unit length with λ > 0. The resulting electric field is \vec{E} = 5 \lambda \frac{x \vec{i} + y \vec{j}}{x^2 + y^2}<br />.

Compute the flux of \vec{E} outward through the cylinder x^2 + y^2 = R^2, for 0 \leq z \leq h.

Homework Equations



d\vec{A} = \vec{n}dA

The Attempt at a Solution



There are three surfaces to compute flux through and I believe that I have to sum those to get the answer. The three surfaces are the two circles that cap the cylinder and the cylindrical face. The unit normal vectors for these are k, -k and \frac{x \vec{i} + y \vec{j}}{\sqrt{x^2 + y^2}}.

My first question:
So for each surface I have an integral over the surface which is the dot product of E and ndA?

My second question:
Because the vector equation of E doesn't involve k and the normal vector equations for the top and bottom don't include i or j, the product E*ndA is 0dA? So those integrals are 0 and I just have to find the integral with the cylindrical surface normal?
 
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smagf said:

Homework Statement



The z-axis carries a constant electric charge density of λ units of charge per unit length with λ > 0. The resulting electric field is \vec{E} = 5 \lambda \frac{x \vec{i} + y \vec{j}}{x^2 + y^2}<br />.

Compute the flux of \vec{E} outward through the cylinder x^2 + y^2 = R^2, for 0 \leq z \leq h.


Homework Equations



d\vec{A} = \vec{n}dA

The Attempt at a Solution



There are three surfaces to compute flux through and I believe that I have to sum those to get the answer. The three surfaces are the two circles that cap the cylinder and the cylindrical face. The unit normal vectors for these are k, -k and \frac{x \vec{i} + y \vec{j}}{\sqrt{x^2 + y^2}}.

My first question:
So for each surface I have an integral over the surface which is the dot product of E and ndA?
Yes.

My second question:
Because the vector equation of E doesn't involve k and the normal vector equations for the top and bottom don't include i or j, the product E*ndA is 0dA? So those integrals are 0 and I just have to find the integral with the cylindrical surface normal?
Yes.

You could also use Gauss's theorem (divergence theorem) to integrate over the volume of the cylinder.
 
welcome to pf!

hi smagf! welcome to pf! :wink:
smagf said:
My first question:
So for each surface I have an integral over the surface which is the dot product of E and ndA?

yes :smile:
My second question:
Because the vector equation of E doesn't involve k and the normal vector equations for the top and bottom don't include i or j, the product E*ndA is 0dA? So those integrals are 0 and I just have to find the integral with the cylindrical surface normal?

yes :smile:

(btw, in this case it's probably easier to change to cylindrical coordinates before integrating, using of course xi + yj = r :wink:)
 

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