What is the electric flux through the hemispherical surface?

Click For Summary
SUMMARY

The electric flux through a hemispherical surface in a uniform electric field depends on the orientation of the field relative to the surface. When the electric field E is parallel to the axis of the hemisphere, the electric flux is calculated using the formula Φ_E = EA cos(θ), where θ is the angle between the electric field and the normal vector to the surface. In the case where E is perpendicular to the axis, the flux simplifies to Φ_E = EA. However, to accurately compute the flux, one must evaluate the surface integral Φ_E = ∫E · n da, where n is the unit vector normal to the surface element da.

PREREQUISITES
  • Understanding of electric flux and Gauss's Law
  • Familiarity with vector calculus and surface integrals
  • Knowledge of electric fields and their properties
  • Basic concepts of coordinate systems in physics
NEXT STEPS
  • Study the application of Gauss's Law in calculating electric flux
  • Learn about surface integrals in vector calculus
  • Explore the relationship between electric fields and potential energy
  • Investigate the effects of different orientations of electric fields on flux calculations
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism or vector calculus will benefit from this discussion.

physics231
Messages
3
Reaction score
0
A uniform field E is parallel to the axis of a hollow hemisphere of radius R. a) What is the electric flux through the hemispherical surface? b) What is the result if E is instead perpendicular to the axis?

Here is what I've interpretted so far. If the field is parallel to the surface, then the electric flux = EA cos(theta). With the angle being 0, I came up with the answer as just EA Therefore that is my answer on part a).

On part b) if the field is perpendicular then the electric flux is just = EA, therefore making that my answer. But I know this isn't right. What step am I missing? Thank you
 
Physics news on Phys.org
Well the field is certainly not parallel to the surface in both cases. You'll have to evaluate the surface integral

[tex]\Phi_E=\int\vec{E}\cdot \hat{n}da[/tex]

where [itex]\hat{n}[/itex] is the unit vector parallel to the surface element da.

Hint: If you chose a coordinate system in which the origin is on the center of the hollow sphere and in which the z-axis IS the axis of the hollow hemisphere, then what are [itex]\vec{E}[/itex] and [itex]\hat{n}[/itex]?
 
Last edited:

Similar threads

Replies
20
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
Replies
6
Views
3K
Replies
26
Views
2K