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## Homework Statement

Find the electrostatic force of interaction between two halves of a spherical conductor of radius R carrying a charge Q.

This thread already exists- https://www.physicsforums.com/threads/electric-force-between-two-halves-of-a-sphere.317380/ but is closed.

## Homework Equations

Electrostatic pressure:

dF/dA = σ

^{2}/2ε

_{0}

## The Attempt at a Solution

I understand how to do it using the method outlined in the above mentioned thread, but I want to do it using the argument stated in the last post of the thread, the one that mentions this at the very end: "By the way, I think the "supposed way" to do this Griffith problem is to use the formula for surface pressure of a conductor earlier in the chapter and add up the integrate z-component of the force on the base and the side of a hemisphere. Be ware, you need to make a correct physical argument about the direction of the pressure/force."

So, to solve it, I tried splitting up the northern hemisphere as elementary rings. Basically, here's what I'm trying to do:

Take an elementary area dA containing a charge dq on an elementary ring of radius r. The force acting on dq is

dF = ( σ

^{2}dA ) / 2ε

_{0}

Now comes my problem-

I'm not sure what the net force on the ring should be, should it be ∫dFsinθ or ∫dFcosθ, i.e. should I take the vertical components or the horizontal components?

So, I tried using both and got the right answer using ∫dFsinθ

Here's what I did:

The net force on the ring is

F (ring) = ∫dFsinθ = ∫( σ

^{2}2πRcosθRdθ sinθ) / 2ε

_{0}

and F(hemisphere) = ∫F(ring)

Integrating from 0 to π/2 I get Q

^{2}/ 32πεR , which is correct.

But I still don't get why I'm choosing ∫dFsinθ over ∫dFcosθ.