Electric flux through hemisphere

Question

My teacher posed this question and it got me thinking;

The electric flux through the curved surface area of a hemisphere of radius R when it is placed in a uniform electric field is?

Before this, I was taught the definition of flux as the number of field lines passing perpendicularly through an area. (If the lines aren't perpendicular, we use the component of field line that is)

Now basically it's like this(not able to attach a diagram): if the hemisphere is the bowl, the field lines are coming perpendicularly into the bowl.

I do realise that only the portion of hemisphere right in front of the circular opening would get all the field lines but the area vector would keep on changing directions all over the surface, which would change the angle between E and A, flux is the dot product of E and A, so flux would (should, at least) get affected but my teacher told me the flux is $EπR^2$ and now I'm confused because just prior to the question, he taught us about how varying angles between E and A affects flux. I looked up an online solution and it matches with my teacher's. Please help.

Answer 1

You're right, the angle between $\mathbf{E}$ and the infinitesimal area $\text{d}\mathbf{A}$ does affect the value of the flux, it's for this reason that the flux isn't $2\pi R^2 E_0$ as you might "naively" imagine ($2\pi R^2$ being the area of a hemisphere).

Here's a simple "intuitive" way to see it: since the field is constant everywhere on the surface, all you need to find is the product of the field magnitude with the projection of the surface on the $xy-$plane (i.e. perpendicular to the direction of the field). Imagine the hemisphere to be placed in front of a wall, and the electric field is a "torchlight" that's shining onto its cross-section. What is the area of the total light that has been blocked? It will be the area of the shadow cast by the sphere, which is just $\pi R^2$ if the light is uniform everywhere. The field flux passing through that area is then just the product of this "projected" area and the field strength, $E_0 \pi R^2$.

If you're not convinced, it's really not hard to actually calculate it; I'd suggest doing it as an exercise. I'll sketch out the procedure for you: The electric flux is given by $$\phi_E = \iint\mathbf{E}\cdot\text{d}\mathbf{A},$$ and in your case $\mathbf{E} = E_0 \mathbf{\hat{z}}$ with $E_0$ being a constant, meaning that $$\phi_E = E_0 \iint\mathbf{\hat{z}}\cdot\text{d}\mathbf{A},$$

                          

You should be able to see from the image above that the area element on the surface of the sphere (called $\text{d}^2\mathbf{S}$ in the image) is $R^2 \sin{\theta}\text{d}\theta \text{d}\phi \mathbf{\hat{r}}$. The important point to realise is (as you pointed out) that $\mathbf{\hat{r}\cdot \hat{z}} =f(\theta)$, where $f(\theta)$ is a very simple function of $\theta$. (I'd urge you to calculate it geometrically.)

Using this fact, you can find that

$$\phi_E = E_0 \int_0^{2\pi} \text{d}\phi \int_0^{\pi/2} R^2 \sin{\theta} f(\theta) = 2 \pi R^2 E_0 \int_0^{\pi/2} \sin{\theta} f(\theta).$$

If you've calculated everything as expected, you should find that $\phi_E = \pi R^2 E_0$.

Answer 2

You are thinking along the right lines (pardon the pun), but the total flux is still $\phi_E = \pi R^2 E$. You are correct that the field lines will be at different angles to the normal vector at different points on the curved surface; if you divided the curved surface up into lots of smaller areas, the flux through each would be $d\phi_E = \textbf{E} \cdot \textbf{dA},$ with the dot product capturing the fact that they are not always 'aligned' with each other. If you added up all the small fluxes over the curved surface area, you would get $$ \int_{\text{Curved Surface}} \textbf{E} \cdot \textbf{dA} = \phi_E = \pi R^2 E$$ if you performed the integration in, say, polar coordinates. (I haven't included any calculations here to keep the answer more readable, but I'm more than happy to include everything if you need. It's only a simple problem in multivariable calculus.)

However, there is a much easier way of getting the same result if you think a little creatively. All the flux that passes through the curved surface of the hemisphere also passes through the flat base. In fact, it does not matter what the shape on the other side is -- whether a hemisphere or a cone or anything else -- just as long as it is a closed surface and the Electric Field is constant, it is going to 'catch' as much flux as the flat base.

Therefore, the total flux is always going to be $\phi_E = E \cdot \text{Area of the Base}.$