A conducting shell kept in a uniform electric field

[rude man]

You've done extremely well! Especially for a high school student!
The notion of electrostatic pressure due to surface charge density is perhaps a bit mysterious. Here is a link giving a good derivation of the pressure formula at a conductor of p = σ²/2ε₀

http://www.physicspages.com/2011/10/31/electrostatic-pressure/

 

[Titan97]

$$dF=\frac{\sigma}{2\epsilon_0}\sigma dA=\frac{9}{2}\epsilon_0E_0^2\cos^2\theta\hat{r}$$
$$dA=2\pi a^2\cos\theta\sin\theta d\theta$$
Only force along $z$ remains (that is, along rcosθ). The other components cancels out.
$$dF=9\epsilon_0E_0^2\pi a^2\cos^3\theta\sin\theta d\theta$$
$$F=\frac{9}{4}\epsilon_0E_0^2\pi a^2$$

 

[TSny]

$$F=\frac{9}{4}\epsilon_0E_0^2\pi a^2$$

I think that's correct. Great work!

 

[rude man]

$$dF=\frac{\sigma}{2\epsilon_0}\sigma dA=\frac{9}{2}\epsilon_0E_0^2\cos^2\theta\hat{r}$$
$$dA=2\pi a^2\cos\theta\sin\theta d\theta$$

I have dA = a² sinθ dθ dφ = 2πa²sinθ dθ.

 

[Titan97]

The radius of the ring is $a\sin\theta$. Its width is $ad\theta$. Yes. The area is $2\pi a^2\sin\theta d\theta$.

Then $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\cos\theta$$
since other component cancels out.

The final answer is same. I think I made some typing error. (I also missed 'dA' in the first line).

 

[rude man]

The radius of the ring is $a\sin\theta$. Its width is $ad\theta$. Yes. The area is $2\pi a^2\sin\theta d\theta$.

Then $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\cos\theta$$
.

Why the cos θ term? Looks to me like it should be sin θ, not cos θ, since the "pull" is maximum at the equator where θ = π/2, and zero at the poles where θ = 0 and π.

So that would make $$dF=\frac{\sigma}{2\epsilon_0}\sigma dA\sin\theta$$, with
dA = 2πa²sin θ dθ.

 

[Titan97]

@rude man

I need to take the cosine component.

 

[rude man]

@rude man
View attachment 99332
I need to take the cosine component.

You're quite right. I had the wrong orientation of the shell w/r/t my coordinate system.