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There are three parallel metal plates. Two boundary plates are fixed and connected with a galvanic battery, maintained a constant potential difference $\Delta \varphi$. Middle plate was initially in contact with one of the end plates. Then, using an insulating handle, it began to move toward the other plate. Ignoring boundary effects, find the field strengths E1, E2 in the gaps between the plates, depending on the varying distance x.

Answer: $$ E_{1} = \frac{\Delta \varphi x}{d^{2}}, \quad E_{2} = \frac{\Delta \varphi (x + d)}{d^{2}}, $$

where d is the distance between boundary plates.

I don't know how the moving plate affect to the measure of field strengths. Can you help me?

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Please don't ask users to solve homework questions..! –  Waffle's Crazy Peanut Sep 23 '12 at 14:41
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I didn't ask to solve homework question. I asked to help me to understand "...how the moving plate affect to the measure of field strengths...", because without understanding this I can't solve the task. You see it, don't you? So, it will be better if you'll help me. –  PhysiXxx Sep 23 '12 at 14:43
    
Ok, Maxim. I apologise for that silly suspicion. Don't get tensed man..! –  Waffle's Crazy Peanut Sep 23 '12 at 18:30
    
Maxim, the thing is, the answer to the question of how the position of the moving plate affects the field - the formulas for $E_1$ and $E_2$ - is the answer to your homework problem. So actually you are asking for the solution, even if perhaps that's not what you meant to do. Could you clarify what exactly it is that you don't understand in trying to find the formulas? –  David Z Sep 24 '12 at 20:29
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closed as too localized by Qmechanic, David Z Sep 24 '12 at 20:30

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2 Answers

up vote 2 down vote accepted

I think the answer is correct as stated in the problem. To analyze:

1) In the initial condition, with the middle plate connected to the left end plate, calculate the surface charge density required on the right surface of the middle plate (via Gauss' law) to produce the posited electric field $\Delta \phi /d$. (The surface charge density on the left side of the middle plate is zero, since it's facing a connected end plate at the same potential.)

2) Now break the connection between the end and middle plates, without moving the plate. The fields and surface charges on the middle plate are unchanged, but charge on the middle plate is now "stranded" there. From here on, it's the charge on the middle plate, not its potential, that's fixed.

3) Now imagine the middle plate moved to some position x. Its (fixed) charge will re-distribute between its two faces to produce electric fields on either side that integrate up to the fixed total potential $\Delta \phi$. That's enough info to solve for the surface charges and electric fields.

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Consider the middle plate at a distance x from the first plate. The amount of charge on all the plates =0 and the amount of charge on single side of the plate is Q. The potential difference between 1 and middle is x*(V1-V2)/d. The potential difference between middle and 2 is (d-x)*(V1-V2)/d. I think its clear up to this part. The field inside a metal is zero. Hence, the field on the right side should be equal to the field on the right side of the middle plate. And its value is same even if you remove the middle metal plate.

E=(V1-V2)/d=E1=E2

Am sorry if i have taken the question wrong. I havent taken the concept of 1 and middle plate together. When the middle plate loses contact with the first one, let the initial charge on 1 and middle (together) is +q on left and -q on right. Consider the same for the left plate +q on right, -q on left. When the middle one is moved, the charges between the 1 and middle will become +2q on the left of the middle and -2q on the right of the 1. The field between 1 and middle is 2*x*(V1-V2)/d. And, the field between the 2 and middle is (d-x)*(V1-V2)/d.

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I think this answer is incorrect. Imagine the middle plate next to an end plate and connected to it via a wire: the field between them is zero. Now break the connection without moving the plate: the fields won't change, so the field between the two plates remains zero, contrary to your answer. The essence is that the charge on the middle plate is not zero. –  Art Brown Sep 24 '12 at 19:52
    
Edited the answer please check it. –  Fr34K Sep 24 '12 at 20:36
    
um, the moderators don't want us to get too specific, so I don't want to go beyond what I already posted. I'll just reiterate that I think the fields included in the question are correct (and you can check your answer by calculating the total voltage drop your fields give, and also by checking that the net charge density on the middle plate doesn't change with x). –  Art Brown Sep 24 '12 at 20:47
    
I have cross checked my answers. And i am pretty sure about it. –  Fr34K Sep 25 '12 at 6:54
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