184_notes:examples:week5_flux_cylinder

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A constant electric field E is directed along the x-axis. A cylinder with radius R and height h is situated in the field so that the bases of the cylinder are parallel to the xz-plane. What is the electric flux through the cylinder?

Facts

  • The cylinder is closed.
  • The cylinder has radius R and height h.
  • The electric field is directed along the x-axis, which is parallel to the bases of the cylinder.

Lacking

  • Φe
  • A of cylinder, or dA pieces.

Approximations & Assumptions

  • The electric field is constant.
  • The electric flux through the cylinder is due only to E.
  • We can approximate infinitesimally small pieces of the cylinder's area as flat.

Representations

  • We represent the electric flux through a flat surface with:

Φ=EA

  • We represent the situation with the following diagram:

Cylinder

Before we dive into the math, let's reason about the nature of the situation. It will be helpful to visualize how the area vector dA would look for different parts of the cylinder's surface. Here is a visual from a couple perspectives, where each arrow represents the direction of dA at its location. Cylinder Area-Vectors Notice that the area vectors on the bases of the cylinder are pointing along the y-axis. Since the electric field is aligned with the x axis, there will be no flux through the top and bottom of the cylinder. The math is right here: Φtop=EAtop=(Eˆx)(πR2ˆy)=0 The same is true for bottom. For completeness, the math is here: \Phi_{\text{bottom}}=\vec{E}\bullet\vec{A}_{\text{bottom}=(E\hat{x})\bullet(\pi R^2(-\hat{y}))=0

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  • Last modified: 2017/09/15 22:31
  • by tallpaul