Force of interaction between caps

[TSny]

Pranav used the known expression for the field inside S due to a total sphere of charge density $\rho_2$: $E=\rho_2r/(3\epsilon_0)$.

The z-component of this field is then $E_z=\rho_2z/(3\epsilon_0)$. Since it only depends on z, you can get the force on the cap by integrating over disks.

$F=\int_h^R E_z (\rho_1\pi r^2) \,dz$ where $r = \sqrt{R^2-z^2}$. I get the same result as Pranav.

 

[Saitama]

Sorry for my indifferent behaviour towards the thread but I have a genuine reason of doing it. I was busy with my exam which was on 6th April and the subsequent days were also busy for me, I had no intention of dumping the thread.

Thank you haruspex for keeping the the thread alive!

Consider two small regions X, Y in A. X₁ exerts a force on Y₀, while Y₁ exerts an equal and opposite force on X₀. Thus, there is no net force between A₀ and A₁.

Yes, I am agreed. Thank you for stating this, I don't think I could have thought of this argument. :)

t follows that the force between A₀ and B₁ is the same as that between A₀ and S₁.

I calculated this in #13 and plugging the values doesn't seem to give me the right answer.

 

[TSny]

Sorry for my indifferent behaviour towards the thread but I have a genuine reason of doing it. I was busy with my exam which was on 6th April and the subsequent days were also busy for me, I had no intention of dumping the thread.

Hope you did well on your exam!

I calculated this in #13 and plugging the values doesn't seem to give me the right answer.

Uh Oh. Well, maybe haruspex can set us straight. Did you get a force of 3.33 N?

 

[Saitama]

Uh Oh. Well, maybe haruspex can set us straight. Did you get a force of 3.33 N?

Woops, looks like I did not correctly input the values in the calculator. Thanks TSny and haruxpex! 3.33 N is the correct answer. :)

 

[haruspex]

Pranav used the known expression for the field inside S due to a total sphere of charge density $\rho_2$: $E=\rho_2r/(3\epsilon_0)$.

Ah, OK. I did it from first principles, but missed a cancellation of π.