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$$\phi(z)=\frac{\sigma}{4\pi\varepsilon_0}\int_{\frac{-a}{2}}^{\frac{a}{2}}\int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{1}{\sqrt{x^2+y^2+z^2}}~dx~dy$$

I'm not sure how to do this integral. For the first integral w.r.t $x$ i tried to substitute $x=\sqrt{y^2+z^2}\sin{\theta}\implies dx=\sqrt{y^2+z^2}\cos\theta~d\theta$.

The integral then becomes:

$$\phi(z)=\frac{\sigma}{4\pi\varepsilon_0}\int_{\frac{-a}{2}}^{\frac{a}{2}}\int_{-\text{?}}^\text{?}1~d\theta~dy$$

But the bounds are $\text{?}=\arcsin{\frac{a}{2\sqrt{y^2+z^2}}}$ since arcsin is an odd function. However this just makes it even harder to solve. So what is the best way to do this integral?

If it helps I will give the context of the question. I am asked to find the strength of the electric field at a height z above the centre of a square sheet with constant charge density $\sigma$ and side lengths $a$.

Michael Hardy
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NormalsNotFar
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    By symmetry we can integrate only over one-eighth of the square bounded by the $x$-axis, the line $x=a/2$, and the diagonal line $y=x$, and then multiply by $8$. $$ x = r\cos\theta $$ $$ = \frac a 2 \text{ if and only if } r = \frac a {2\cos\theta} = \frac a 2 \sec\theta. $$ Then we have${},\ldots$ $$\qquad$$ – Michael Hardy Feb 01 '16 at 17:39
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    $$ \int_0^{\pi/4} \left(\int_0^{(a/2)\sec\theta} \frac 1 {\sqrt{r^2+z^2}} r,dr \right) , d\theta $$ $$ = \int_0^{\pi/4} \left( \int_{z^2}^{z^2 + (a^2/4)\sec^2\theta} \frac 1 {\sqrt u} \left( \frac 1 2 , du \right) \right) , d\theta $$ $$ = \int_0^{\pi/4} \left( \sqrt{z^2 + \frac{a^2}4 \sec^2\theta} - z \right) , d\theta $$ $$ = \frac{-\pi z} 4 + \int_0^{\pi/4} \sqrt{z^2 + \frac{a^2}4 \sec^2\theta} \ d\theta. $$ (Since this is incomplete I'm posting this as a comment rather than as an answer.) $\qquad$ – Michael Hardy Feb 01 '16 at 17:40
  • u might have a look at my answer here, u cana proceed along the same lines: http://math.stackexchange.com/questions/1122135/average-distance-to-a-random-point-in-a-rectangle-from-an-arbitrary-point/1122286#1122286 – tired Feb 01 '16 at 18:43
  • @tired It turns out that the electric field along the polar axis is a bit easier to compute in Cartesian form than the potential. - Mark – Mark Viola Feb 01 '16 at 19:19

1 Answers1

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By symmetry, the electric field along the $z$ axis will have only a $z$ component with

$$E_z(0,0,z)=\left.-\frac{\partial \phi(x,y,z)}{\partial z}\right|_{(0,0,z)}$$

Therefore, we have

$$E_z(0,0,z)=\frac{\sigma z}{\pi \epsilon_0}\int_0^{a/2}\int_0^{a/2}\frac{1}{(x^2+y^2+z^2)^{3/2}}\,dx\,dy$$

Rather than attempt a transformation to cylindrical coordinates, we proceed here with integrating directly in Cartesian coordinates.

We evaluate the inner integral by making the substitution $x=\sqrt{y^2+z^2}\tan \theta$. This yields

$$\begin{align} \int_0^{a/2}\frac{1}{(x^2+y^2+z^2)^{3/2}}\,dx&=\left.\left(\frac{x}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}\right)\right|_{0}^{a/2}\\\\ &=\frac{a/2}{(y^2+z^2)\sqrt{(a/2)^2+y^2+z^2}} \end{align}$$

Therefore, we have reduced the expression for the electric field along the $z$ axis to

$$E_z(0,0,z)=\frac{\sigma z}{\pi \epsilon_0}\int_0^{a/2}\frac{a/2}{(y^2+z^2)\sqrt{(a/2)^2+y^2+z^2}}\,dy$$

To evaluate the remaining integral, we make the standard trigonometric substitution $y=\sqrt{(a/2)^2+z^2}\tan(u)$. Then, we have

$$\begin{align} E_z(0,0,z)&=\frac{\sigma z(a/2)}{\pi \epsilon_0}\int_0^{\arctan\left(\frac{a/2}{\sqrt{(a/2)^2+z^2}}\right)}\,\,\frac{\cos(u)}{z^2+(a/2)^2\sin^2(u)}\,du\\\\ &=\frac{\sigma z(a/2)}{\pi \epsilon_0}\int_0^{(a/2)/\sqrt{(a/2)^2+(a/2)^2+z^2}} \frac{1}{z^2+(a/2)^2v^2}\,dv\\\\ &=\frac{\sigma }{\pi \epsilon_0}\arctan\left(\frac{(a/2)^2}{z\sqrt{2(a/2)^2+z^2}}\right) \end{align}$$

NOTE 1:

We can recover, of course, the potential along the $z$ axis by integrating the electric field. In this problem, integration by parts facilitates. This is left as an exercise for the reader.

NOTE 2:

As $a\to \infty$, the arctangent goes to $\pi/2\,\text{sgn}(z)$, and we recover the familiar result of the electric field from a uniform surface charge on an infinite surface, namely $\vec E(0,0,0^{\pm})=\pm \hat z\,\frac{\sigma}{2\epsilon_0}$.

NOTE 3:

As $z\to 0^{\pm}$, the arctangent goes to $\pi/2 \,\text{sgn}(z)$ and the electric field is $\vec E(0,0,0^{\pm})=\pm \hat z\,\frac{\sigma}{2\epsilon_0}$

NOTE 4:

As $z\to \pm \infty$, the arctangent goes to $\frac{(a/2)^2}{z^2}\,\text{sgn}(z)$ and the electric field is $\vec E(0,0,z\to \pm \infty)=\pm \hat z\,\frac{\sigma a^2}{4\pi \epsilon_0\,z^2}$, which appears as the field from a point charge $q=\sigma a^2$.

Mark Viola
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  • So calculating the electric field first in such setups is equivalent to feynman integration...interesting something new to think about :) (+1) – tired Feb 01 '16 at 19:58
  • @tired In this case, it does work that way, although we could have gone after the potential directly - it is a bit uglier, I believe, to do it that way. So, Richard's old trick did help a bit ... that along with the OP's stating that it was the electric field sought all along. - Mark – Mark Viola Feb 01 '16 at 20:09
  • Hi Dr.MV,

    Thank you for the detailed response. I am a little lost on how you computed the first (inner) intergral. After using the substitution you suggest I get:

    $$\int_0^{a/2}\frac{1}{(x^2+y^2+z^2)^{3/2}}~dx=\int_{-?}^?\frac{\cos{\theta}}{y^{2}+z^2}~d\theta$$

    Where the $?=\arctan{\frac{a}{2\sqrt{y^2+z^2}}}$

    What am I missing?

     – NormalsNotFar Feb 01 '16 at 20:31
  • In my second integral above, it should be a y^2 in the denominator.. for some reason it is not rendering. – NormalsNotFar Feb 01 '16 at 20:34
  • Dont worry I can see now :) – NormalsNotFar Feb 01 '16 at 21:14
  • You're welcome. My pleasure. And very pleased to hear that you've got it now! - Mark – Mark Viola Feb 01 '16 at 22:54
  • One more thing Dr.MV, I can understand how you get your answer. But the answer given in Griffiths textbook is different to yours, i have been trying to make it look the same through inverse trig identities for the past 2 hours but no luck. Here is Griffiths answer:

    $$\frac{\sigma}{2 \epsilon_0}(\frac{4}{\pi}\arctan{\sqrt{1+\frac{a^2}{2z^2}}}-1)$$

    I might start a new thread for this, though.

     – NormalsNotFar Feb 01 '16 at 23:48
  • Use $\arctan (1)=\pi/4$ and the addition angle formula for the tangent to show equivalence of the expressions. ;-)) – Mark Viola Feb 02 '16 at 00:01
  • Hmmm, I tried starting with griffiths expression, and used the fact that arctan(1)=pi/4 and the double angle formula for arc tans but it still looks like a mess, I am stuck at this point:

    $$\frac{\sigma}{2 \epsilon_0}(\frac{4}{\pi}\arctan{\sqrt{1+\frac{a^2}{2z^2}}}-1)=\frac{2 \sigma}{\pi \epsilon_0}(\arctan{\sqrt{1+\frac{a^2}{2z^2}}}+\arctan(-1))$$ $$=\frac{2 \sigma}{\pi \epsilon_0}\arctan{\frac{\sqrt{1+\frac{a^2}{2z^2}}-1}{\sqrt{1+\frac{a^2}{2z^2}}+1}}$$

     – NormalsNotFar Feb 02 '16 at 00:46
  • You're almost there. Now use the identity $2\arctan(z)=\arctan\left(\frac{2z}{1-z^2}\right)$ and simplify ... you should recover the answer I provided in this post. ;-)) ... Mark – Mark Viola Feb 02 '16 at 03:27
  • Oh my, thank God that is over. You are a life save - Thanks a lot :) – NormalsNotFar Feb 02 '16 at 08:08
  • You're welcome. My pleasure – Mark Viola Feb 02 '16 at 14:51
  • One noteworthy difference between the two answers is that the one I derived is valid for all $z\ne 0$ whereas Griffith's is only valid for $z>0$ – Mark Viola Feb 02 '16 at 15:08