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I have two parallel non conducting charged planes with opposite charges $6\mu C/m^2$, area $A = 3m^2$ and distance between the planes $d = 0.004 m$. I know the potential between these two planes is $$V = \frac{\sigma\times d}{\epsilon_0}$$ But if I put a conducting sheet of thickness $h = 0.001 m$ with the same area $A = 3m^2$ between these two planes, the new potential between the the planes would be

$$V = (2\times\frac{\sigma}{2\epsilon_0} + 2\times\frac{2\sigma}{2\epsilon_0})\times d$$

Is this right? or Am I wrong somewhere? I mounted this schema.

Is that right?

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Hi Joao. I think this slightly disagrees with our policy... – Waffle's Crazy Peanut Mar 24 '13 at 16:49
How can I improve the question? I tried to explain what I did with the image I posted. – Mr. Felix Mar 24 '13 at 16:52
That's why I mentioned slightly. Image isn't gonna provide you the answer anyway. When you try to explain what you've done, there will be no question like "How can I calculate..." It'd be actually, "I did like this... Where am I getting wrong?" <-- sorta something like that. I don't know. I don't make any decision. Maybe, someone could take a look ;-) – Waffle's Crazy Peanut Mar 24 '13 at 16:57
I posted a solution I've tried and I will do some changes like "How can I calculate..." – Mr. Felix Mar 24 '13 at 17:01
I think it's OK now. Hope you like the formatting ;-) – Waffle's Crazy Peanut Mar 24 '13 at 17:08
up vote 2 down vote accepted

The charges on the conducting sheet will be $±\sigma$, not $±2\sigma$. Also, you will have to subtract the thickness of the conducting sheet, $h$ because $E$ is zero there.

So, your final equation should be $$V=(\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0})\times(d-h)$$

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Dude, you shine!! – Mr. Felix Mar 25 '13 at 21:33

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