How Is Force Calculated Between Charged Parallel Plates?

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The discussion revolves around calculating the force between charged parallel plates with charges Q and -Q. The initial calculations lead to the expression F = Q²/(Aε₀), but there is confusion regarding the dependency on distance x. Participants express uncertainty about whether the problem is correctly understood, suggesting that the force F should be an external applied force rather than derived from the given information. There is also mention of integrating forces due to differential charges on the plates, but the complexity of this approach raises further questions. Ultimately, the participants feel stuck and uncertain about the accuracy of their solution and the clarity of the problem statement.
Hannisch
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Homework Statement


Two parallel plates with area A and charges Q and -Q are being pulled apart by force F, so that the distance between, x, slowly increases. Express F in terms of Q, A and x.

Homework Equations


-\Delta PE=W

W=\int Fdx

\Delta PE=q_{0}\Delta V

Q=C\Delta V

C=\frac{A\epsilon_{0}}{d}

The Attempt at a Solution


Q=C\Delta V

Q=\frac{A\epsilon_{0}}{x}\Delta V

\Delta V=\frac{Qx}{A\epsilon_{0}}

q_{0}\Delta V=\Delta PE and in this case q0= -Q

\Delta PE=\frac{Qx}{A\epsilon_{0}} (-Q)

W=-\Delta PE=-\frac{-QQx}{A\epsilon_{0}}

W=\frac{Q^{2}x}{A\epsilon_{0}}

W=\frac{Q^{2}x}{A\epsilon_{0}}=\int Fdx

Force is then equal to the derivative of my expression with respect to x.

F=\frac{Q^{2}}{A\epsilon_{0}}How can I get an answer independent of x? It.. just doesn't make sense to me. My friend and I both got this answer independently. Where have we gone wrong, or is it correct or.. something? I'm rather confused.

(Ohhh the pain, did you know that Ctrl+W closes a tab [at least in Firefox]? Yeah, I was going to write W and managed to hit Ctrl instead of Shift.. towards the end of my attempt at a solution.)
 
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I don't understand the question. The force F "pulling the plates apart" must be an external applied force and there is no way to calculate it from the information given. Perhaps you are supposed to find the force of attraction between the oppositely charged plates. That would be ROUGHLY F = kQQ/x^2. The question does say "express in terms of Q, A and x" which means it is okay to have an x in the answer.

I see you know about integration. That means you could do the force of attraction between the plates more accurately by considering a tiny charge dq in a bit of the plate area dA and the force it feels do to all the other dQ's on the other plate - integrating over dQ and dq to get the total force. A double integral. Oh, I don't think you can do all that without knowing the shape of the plate. The shape will matter.

Sorry - I either don't know how to do it or there isn't enough information given.
 
Ah, there was a bit more text in the question from the beginning, but there really wasn't anything important (it was on a test and I can't seem to remember the exact wording, however what I wrote was the useful part). He provided us with a little hand made drawing too, of like a parallel plate capacitor with a force F on the right plate ..

Probably I've understood the question wrong or something.. but how on Earth can both my friend and I get the same answer? And another friend of mine (who said she got Q^2x^2/Aepsilon, but then she realized she did the primitive of it instead of the derivative, so yeah..).

And I bet you my teacher translated a question from Swedish to English. Wouldn't surprise me at all.. and he's not thet best in English. I think I'll just have to give up on this question then. Bummer, it's been nagging at me. (And a double integral doesn't really make sense to me, but then again I've only dealt with integrals from a textbook in Swedish, you never know.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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