Average Force

Consider two points on the square: $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between the points are:

$D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ Average distance is:

$D_{av} = \int_{0}^{1}\int_{0}^{1} \int_{0}^{1} \int_{0}^{1}\left(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\right) dx_1 \ dy_1 \ dx_2 \ dy_2$

This is difficult to solve numerically. Fortunately, this problem of average distance is a popular one and its solution is available on the internet.

$D_{av} \approx 0.5214$

Using this expected distance, the expected force can be computed.

$F = \frac{q^2}{4 \pi \epsilon_o D_{av}^2}$

$\therefore F \approx \frac{3.678 \ q^2}{4 \pi \epsilon_o}$

I am not sure if this method is correct, however. This approach was a trial and it turned out to be correct. The rigorous way of solving for the expected force is:

$F = \frac{q^2}{4 \pi \epsilon_o} \int_{0}^{1}\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \left(\frac{1}{D^2}\right) dx_1 \ dy_1 \ dx_2 \ dy_2$

I do not know if this method gives the same result as solving a quadruple integral is computationally expensive. There might be a way to get a closed-form solution, but I have not attempted to solve the integral exactly as yet. I am sceptical about the correctness of my solution.

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Is it correct if the following can be proved:

$\int_{0}^{1}\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \left(\frac{1}{D^2}\right) dx_1 \ dy_1 \ dx_2 \ dy_2 = \frac{1}{D_{av}^2}$

And I am not sure how to do this.