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This is the original question:

The space between two parallel infinite planes $ x=0 $ and $ x=L $ is filled with charge of the density $ \rho\left(x\right)=ax^{2}\left(L-x\right) $ where $ a $ is a constant. There are no other charges and no external uniform field.

Find the electric field in the whole space.

Now, I want to use Gauss's law to calculate the electric field. for $ x>L $ I chose a cylinder which its base is above the field (as you can see in the image below): Assume the cylinder's base area is <span class=$ A $" />

Now we have from Gauss's law we have

$ E\cdot2A=\frac{1}{\varepsilon_{0}}\intop_{0}^{L}ax^{2}\left(L-x\right)dx\cdot ds $

$ E\cdot2A=\frac{A}{\varepsilon_{0}}\intop_{0}^{L}ax^{2}\left(L-X\right) $

And thus $ E=\frac{a}{2\varepsilon_{0}}\left(\frac{Lx^{3}}{3}-\frac{x^{4}}{4}\right) $

Now I want to find the electric field inside the layer. So I want to chose a cylinder as you can see in the image: enter image description here

And I want to use Gauss's law again and write :

$ E\cdot2A=\frac{1}{\varepsilon_{0}}\intop_{?}^{?}\rho\left(x\right)dV $

But I'm not sure how to choose the upper and lower bound of the integral, in order to calculate the charge inside. The right answer according to my book is

$ E=\frac{a}{2\varepsilon_{0}}\left(\frac{Lx^{3}}{3}-\frac{x^{4}}{4}\right) $

And I get this result only if I'm choosing the lower bound to be $ 0 $ and the upper bound to be $ x $. But I do not understand why we can choose the lower bound to be $ 0 $ isnt lower base of the cylinder should be inside the layer as well as the upper base? and why I cannot get this result for a diffrent choice of upper and lower bounds?

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Since, you have been asked to calculate electric field between the plates (not between two interior points), you can't just choose two arbitrary points in between those plates and find the electric field. You have to choose either one of the plates as a reference point as to from which you calculate the electric field to an arbitrary point $x$ measured from the plate you choose to the other plate.

Now, you have to decide which of the plate to choose from either one. But there is a constraint here. You have to necessarily choose the direction you have been given. And the given direction is the one in $'x'$ mentioned in the problem is increasing.

The lower base at $x=0$ is neither inside nor outside. It forms the boundary between both the interfaces and hence, this being a boundary value is fit for reference. Also, note that even though $x=0$ is a boundary surface, the electric field emanating from this point is well between those plates. Hope that helps.

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  • $\begingroup$ If I want to calculate the electric field outside of the layer, I chose a cylinder such that both of the bases are outside the layer. Why cant I chose a cylinder such that just one of the bases would be outside in this case as well? $\endgroup$ – euler 0.0 Jul 30 '20 at 15:48
  • $\begingroup$ @euler0.0 You could just do that. Only thing you need to ensure is that you are not missing out on any electric charge. $\endgroup$ – Rounak Jul 30 '20 at 16:00
  • $\begingroup$ @euler0.0 A thing which I glanced over earlier is that (calculation after the 1st pic) after the integration you haven't put limit $x=L$. When you do that, you will see that choosing cylinders of varied length doesn't affect the physical result (as long as both the boundary surfaces are included). $\endgroup$ – Rounak Jul 30 '20 at 16:09
  • $\begingroup$ I see. Thanks a lot, still I can understand why we cannot choose the cylinder totally in the layer in order to find the electric field inside the layer. I have seen other examples where the layer were from $ z=-d $ to $ z=d $, meaning the length of the layer was $ 2d $ and the lecturer chose a cylinder which its upper base coordinate was $ z $ and the lower base coordinate was $ -z $, for $ z<d $, and thus the integral bounderies were $ -z,z $. But I will read your comments again and think about it. Thanks. $\endgroup$ – euler 0.0 Jul 30 '20 at 16:23
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For this problem it is easiest use the differential version of Gauss law: $$ \frac{\partial E_x}{\partial x}= ax^2(L-x) ,\quad 0<x<L $$ and $$ \frac{\partial E_x}{\partial x}=0, \quad \hbox{elsewhere} $$

Therefore:

i) $E_x$ is constant outside the charge layer, but as there is "no external field" the fields must be equal and opposite in the regions above and below the charge layer.

(Your expression for the field outside the layer still constins $x$ and is therefore wrong.)

ii) $E_x = a\left(\frac {1} {3} x^3L - \frac 14 x^4\right)+C$ inside the charge layer.

iii) Choose the consatant $C$ to satisfy (i) and you are done.

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  • $\begingroup$ I see. According to your answer even chosing the bounderies of the integral to be $ 0,x $ wouldnt get the right result, (because the constand would be missing, and obviously the constant wouldn't be 0. It would be $ \frac{-\pi kaL^{4}}{6} $. Thanks, your way is indeed easier. Can you explain how you would solve it in the way I suggested ? Is there a right choice for the integral bounds that would get me the right result you presented? $\endgroup$ – euler 0.0 Jul 30 '20 at 17:13

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