What is the electric flux through the hemispherical surface?

[physics231]

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

 

[quasar987]

Well the field is certainly not parallel to the surface in both cases. You'll have to evaluate the surface integral

$$\Phi_E=\int\vec{E}\cdot \hat{n}da$$

where $\hat{n}$ 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 $\vec{E}$ and $\hat{n}$?

 

[kyle8921]

for your help.

Your interpretation for part a) is correct. The electric flux through the hemispherical surface would indeed be EA, since the angle between the electric field and the surface is 0 degrees.

For part b), you are correct that the electric flux would still be EA. However, the angle between the electric field and the surface is now 90 degrees, not 0 degrees. This means that the cosine of 90 degrees is 0, so the electric flux would be 0. This makes sense intuitively, as the electric field is now perpendicular to the surface, so no electric flux would pass through it.

In summary, your answers for both parts a) and b) are correct. Just make sure to consider the angle between the electric field and the surface when calculating the electric flux.