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Potential at different plate

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In the solution they have done $V_2 -V_1 = V_2 -V_3$

But I want to know can we solve it by taking $$V_1 -V_3=0$$ $$V_1 = V_3$$
from that we will get $$\frac{-x(3d)}{A\epsilon}= \frac{(x-5Q)(3d)}{A\epsilon}$$ $$ -x=5Q+x$$
But using that I am not getting the answer .

asked Mar 27, 2017 in Physics Problems by koolman (4,116 points)
edited Mar 27, 2017 by koolman
Please show your work from $V_1=V_3$ to the answer you get.
1. You are making the same mistake as the given solution. The electric field between 1& 2 is not $\frac{\sigma}{\epsilon}$. It is $\frac{2\sigma}{\epsilon}$ because there is the same contribution from both plates, which carry equal and opposite charges. However, this does not make any difference to the answer, neither in your attempt nor in the given solution.

2. Your calculation assumes that the electric field is the same between 1 & 2 as between 2 & 3. This is not correct. You are ignoring the contributions from the charges on the inner plate.
It is $\frac{\sigma}{\epsilon}$ as given here http://hyperphysics.phyastr.gsu.edu/hbase/electric/elesht.html

That's what I am confused , how to consider  the contribution of inner charge.
My mistake. My point 1. is incorrect. The field is $\frac{\sigma}{\epsilon}$ as in [What is the electric field in a parallel plate capacitor?](http://physics.stackexchange.com/q/65191)
The book solution shows how to  consider the contribution of the inner charge. Is there some reason why you don't like the book solution? Do you think it is wrong?
I am happy with the book solution . I just want to know if is there any other method to solve Or what is the mistake is my method ?
I have told you what I think is wrong with your solution in my point #2 above. If you want to discuss further please explain what you think you are doing in your solution. Why do you think that what you have written is correct?  

I cannot think of an alternative method.
The electric field due to plate 1 is $\frac{-x}{A\epsilon}$

Hence The potential of plate1 is $\frac{-x(3d)}{A\epsilon}$ (V=E.d)

Similarly Potential of plate 3 is  $ \frac{(x-5Q)(3d)}{A\epsilon}$

Potential of plate 1& 3 are same . So I have equated them .
And find the value of x.
Line 1 is ok.  

What you are calculating in lines 2 and 3 is not potential but potential difference between 1 and 3. Potentials are always arbitrary, although "earth" is usually called zero potential. Plates 1 and 3 are connected to earth, so their potentials are zero.

You are saying that the PD from 1 to 3 equals minus the PD from 3 to 1. This will always be true whatever the charges on 1 and 3.  

Moreover, as I have pointed out in #2 above, you are ignoring the electric fields due to charges on plate 2. You are assuming that the field at the surface of 1 is the only electric field between 1 and 3, and the same for the field at the surface of 3. You are assuming that the field in the region 1-2 is the same as in the region 2-3 : this is not necessarily correct.
Thanks I got it .

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