- #1

Hamal_Arietis

- 156

- 15

- 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
- [tex] σdS=σdxdy[/tex]

[tex] V=\int \frac{\sigma dS}{r}[/tex]

I found out the equation of electric potential, that is

[tex]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}}[/tex]

but I couldn't calculate the integral.

It seems convenient if we use the polar coordinate, that is, assume that

[tex]x=rcos\theta, y=rsin\theta[/tex]

then

[tex]\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}}[/tex]

But I find it difficult to calculate.

I also think another method that is find the electric field $$\vec E$$ then

[tex] V=\int \vec E \vec dl[/tex]

but it seems hard to integrate too .

[tex]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}}[/tex]

but I couldn't calculate the integral.

It seems convenient if we use the polar coordinate, that is, assume that

[tex]x=rcos\theta, y=rsin\theta[/tex]

then

[tex]\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}}[/tex]

But I find it difficult to calculate.

I also think another method that is find the electric field $$\vec E$$ then

[tex] V=\int \vec E \vec dl[/tex]

but it seems hard to integrate too .

Last edited: