Potential of a square plate

[Hamal_Arietis]

Homework Statement

Find the electric potential at a height d above the center of a square sheet
(side 2a) carrying a uniform surface charge σ.

Relevant Equations

$$σdS=σdxdy$$
$$V=\int \frac{\sigma dS}{r}$$

I found out the equation of electric potential, that is
$$V=\int_{-a}^a \int_{-a}^{a} \frac{σdxdy}{4 \pi \epsilon_0\sqrt{x^2+y^2+d^2}}=\int_{0}^a \int_{0}^{a} \frac{σdxdy}{\pi \epsilon_0\sqrt{x^2+y^2+d^2}}$$

but I couldn't calculate the integral.

It seems convenient if we use the polar coordinate, that is, assume that
$$x=rcos\theta, y=rsin\theta$$
then
$$\frac{\pi\epsilon_0 V}{\sigma}=\int_{0}^{\pi/4} \int_{0}^{\frac{a}{cos\theta}} \frac{rdrd\theta}{\sqrt{r^2+d^2}}+\int_{\pi/4}^{\pi/2} \int_{0}^{\frac{a}{sin\theta}} \frac{rdrd\theta}{\sqrt{r^2+d^2}}$$

But I find it difficult to calculate.
I also think another method that is find the electric field $$\vec E$$ then
$$V=\int \vec E \vec dl$$

but it seems hard to integrate too .

 

[haruspex]

Perhaps the double integral is obscuring it.
Could you solve $\int\frac{dx}{\sqrt{x^2+c^2}}$?

 

[Hamal_Arietis]

Perhaps the double integral is obscuring it.
Could you solve $\int\frac{dx}{\sqrt{x^2+d^2}}$?

$$\int\frac{dx}{\sqrt{x^2+d^2}}=log |x+\sqrt{x^2+d^2}|+C$$

but my fomular is
$$ \int\frac{rdr}{\sqrt{r^2+d^2}}=\sqrt{r^2+d^2} +C$$

the problem is when I replace $r=\frac{a}{cos\theta}$, the integral became more complicated.
the double integral became
$$\int_0^{\frac{\pi}4} (\sqrt{\frac{a^2}{cos^2\theta}+d^2}-d)d\theta$$

 

[Delta2]

I don't think its good idea to involve polar coordinates because the region of integration is a square $[0,a]\times[0,a]$ in the plane z=0 (xy plane). So i believe cartesian coordinates are just fine.

However when i tried this integral at wolfram i get a not so simple expression
https://www.wolframalpha.com/input/?i=integral+1/sqrt(x^2+y^2+d^2)dxdy

The first step (integration with respect to x or y) should be easy and I see haruspex post is toward this direction.

$$\int \frac{dx}{\sqrt{x^2+y^2+d^2}}=\ln {|x+\sqrt{x^2+y^2+d^2}|}+c=f(y)$$

The second step $$\int f(y) dy$$ is where I guess difficulties rise and it doesn't have a nice simplified expression.

 

[vela]

Have you solved for the potential a distance $d$ above the center of a line of charge of length $L$? It's a pretty commonly covered problem in intro physics. You can use that result and reduce this problem to a one-dimensional integral.

 

[haruspex]

Have you solved for the potential a distance $d$ above the center of a line of charge of length $L$? It's a pretty commonly covered problem in intro physics. You can use that result and reduce this problem to a one-dimensional integral.

Hasn't that been achieved in the first line of post #3?

 

[Hamal_Arietis]

I tried it. And I found the answer. by considering $d=d_0$ as the potential origin. The potential is
$$V=\frac{\sigma}{\pi\epsilon_0}(a\log{(\frac{d_0^2+a^2}{d^2+a^2}})^\frac{1}{2}+d_0\tan^{-1}(\frac{a}{d_0})-d\tan^{-1}(\frac{a}{d}))$$

It seems correct because when I calculate the limit of V when $a$ into infinite, then
$$V=\frac{\sigma}{2\epsilon_0}(d_0-d)$$

Thanks for helping

Have you solved for the potential a distance $d$ above the center of a line of charge of length $L$? It's a pretty commonly covered problem in intro physics. You can use that result and reduce this problem to a one-dimensional integral.