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.
Best Answer
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 $\phi_E = \pi R^2 E_0$.