Placing charges on a plate

Suppose that the plate is $a$ m square. The electrostatic potential $\varphi_O$ at the centre of the square plate, due to the charge distribution on the plate, is $\varphi_O \; = \; \frac{\sigma}{4\pi \varepsilon_0 (2a)^2}\int_{-a}^a \int_{-a}^a \frac{dx\,dy}{\sqrt{x^2+y^2}} \; = \; \frac{\sigma}{4\pi \varepsilon_0 a^2}J(a)$ while the electrostatic potential at one of the corners of the plate is $\varphi_C \; = \; \frac{\sigma}{4\pi \varepsilon_0(2a)^2}\int_0^{2a}\int_0^{2a}\frac{dx\,dy}{\sqrt{x^2+y^2}} \; = \; \frac{\sigma}{16\pi \varepsilon_0a^2}J(2a)$ where $J(b) \; = \; \int_0^b \int_0^b \frac{dx\,dy}{\sqrt{x^2+y^2}}$ Changing variables makes it clear that $J(b) \,=\, bJ(1)$ for all $b > 0$ . In fact, $J(1) = 2\sinh^{-1}1$ . Thus $\varphi_O \; = \; \frac{\sigma}{4\pi\varepsilon_0a}J(1) \qquad \varphi_C \; = \; \frac{\sigma}{8\pi\varepsilon_0a}J(1)$ and hence $\varphi_C = \tfrac{1}{2}\varphi_O$ . The minimum work done bringing in a test particle $q$ from the point $A$ at infinity to one of the corners of the square is $W_1 = q\varphi_C = 1$ J. The minimum work done bringing in a test particle $q$ from $A$ to the centre of the square is $W_2 = q\varphi_O = 2W_1 = 2$ J.

Use dimensional analysis, we can easily obtain that $W_1$ has form $\frac{k q\sigma a}{\varepsilon_0}$ , where a is the side of the square and k is constant coefficient.

Imagine that the center of the plate is also the corners of 4 smaller squares, whose side is $\frac{a}{2}$ .

Therefore, $W_2=4*\frac{k q\sigma a}{2\varepsilon_0}=2W_1=2(J)$ .

Note: the first time I solved this problem, I didn't notice the correlation between two position and had to calculate the coefficient $k=\frac{ln(1+\sqrt{2})}{2\pi}$ . The coefficient in $W_2$ is two times bigger.

Let us begin by assuming that the expression for potential at the center of a charged plate exists as a function of sigma and L, where L = side of the square plate, and sigma the charge density.

By dimensional analysis, P varies as Sigma x L . So for a corner, we have 1/4th of what would result for Sigma and 2L. For a center, therefore, this yields twice the potential of the corner. Hence 2 J.

That problem has the same formulation, which is W1 and W2 are directly proportional. So, to equalize the W1 (1J), we need 2J because it needs twice to move the same charge from point A to the center of the plate.