# How Do You Calculate the Electric Potential of a Square Plate?

• Hamal_Arietis
In summary: Yes, I have solved for the potential a distance ##d## above the center of a line of charge of length ##L##.
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 .

Last edited:
Delta2
Perhaps the double integral is obscuring it.
Could you solve ##\int\frac{dx}{\sqrt{x^2+c^2}}##?

Hamal_Arietis
haruspex said:
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
$$\int_0^{\frac{\pi}4} (\sqrt{\frac{a^2}{cos^2\theta}+d^2}-d)d\theta$$

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.

Hamal_Arietis
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.

Hamal_Arietis
vela said:
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 and Delta2
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
vela said:
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.

Delta2

## 1. What is the potential of a square plate?

The potential of a square plate refers to the amount of electrical potential energy that is present at any point on the surface of the plate. It is a measure of the work required to move a unit charge from infinity to that point on the plate.

## 2. How is the potential of a square plate calculated?

The potential of a square plate can be calculated by using the equation V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge on the plate, and r is the distance from the point on the plate to the center of the plate.

## 3. What factors affect the potential of a square plate?

The potential of a square plate is affected by the charge on the plate, the distance from the point on the plate to the center of the plate, and the Coulomb constant. It is also affected by the presence of other charges or conductors in the surrounding area.

## 4. How does the potential of a square plate change with distance?

The potential of a square plate follows an inverse relationship with distance. This means that as the distance from the point on the plate to the center of the plate increases, the potential decreases. This is because the electric field strength decreases with distance.

## 5. What is the difference between potential and potential energy of a square plate?

Potential refers to the amount of electrical potential energy present at a point on the surface of the plate, while potential energy refers to the total amount of energy that a charge possesses due to its position in an electric field. The potential energy of a square plate is directly proportional to the potential, but it also takes into account the charge of the plate and the distance from the plate to a reference point.

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