Method of image charges (part II)

The charge and three image charges are symmetrical arranged around the origin in an rectangle with alternating signs. The charges can be grouped into two dipoles with a common perpendicular bisector at $x = 0$ or $y = 0$ such that $\phi (0, y) = \phi (x, 0) = 0$ and the metal surface is an equipotential surface as required.

All four charges together create a potential $\begin{aligned} \phi(x,y) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i = 0}^3 \frac{q_i}{|\vec r - \vec r_i|} \\ &= \frac{q}{4 \pi \varepsilon_0} \sum_{n = \pm 1} \sum_{m = \pm 1} \frac{n m}{\sqrt{(x - n a)^2 + (y - m b)^2}} \end{aligned}$ where $\vec r_0 = (a,b)$ is the position vector of the actual charge. You can easily calculate that $\phi(0,y) = \phi(x,0) = 0$ .

The electric field $\vec E(x,y)$ of the image charges results $\vec E(x,y) = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{\left( \begin{array}{c} x + a \\ y + b \end{array}\right) }{[(x + a)^2 + (y + b)^2]^{3/2}} - \frac{\left( \begin{array}{c} x - a \\ y + b \end{array}\right) }{[(x - a)^2 + (y + b)^2]^{3/2}} - \frac{\left( \begin{array}{c} x + a \\ y - b \end{array}\right) }{[(x + a)^2 + (y - b)^2]^{3/2}} \right]$ Thus, the charge $q$ at $\vec r = (x,y) = (a,b)$ experiences a force $\begin{aligned} \vec F &= q \vec E(a,b) \\ &= \frac{q^2}{4 \pi \varepsilon_0} \left[ \frac{1}{8 [a^2 + b^2]^{3/2}} \left( \begin{array}{c} 2a \\ 2b \end{array}\right) - \frac{1}{8 b^3} \left( \begin{array}{c} 0 \\ 2b \end{array}\right) - \frac{1 }{8 a^3} \left( \begin{array}{c} 2a \\ 0 \end{array}\right) \right] \\ &= -\frac{q^2}{16 \pi \varepsilon_0} \left( \begin{array}{c} 1/a^2 - a/[a^2 + b^2]^{3/2} \\ 1/b^2 - b/[a^2 + b^2]^{3/2} \end{array} \right) \end{aligned}$ If we plug in the numbers, we get the final result $\begin{aligned} \vec F &\approx \left( \begin{array}{c} - 0.199 \\ - 0.023 \end{array} \right) \,\text{N} \\ |\vec F| &= \sqrt{F_x^2 + F_y^2} \approx 0.2 \,\text{N} \end{aligned}$

By the method of the image charges, the following image charge arrangement can be made

In the picture above the red dots represent positive charge and the blue dots represent the negative and all charges have equal magnitude of $q = 1 \, \mathrm {\mu C}$

Now from the Coulomb's law we have(as shown in the above picture):

$\large \vec{F_1}=\frac{-q^2}{4\pi\epsilon_0(2a)^2}a_y$

$\large \vec{F_2}=\frac{-q^2}{4\pi\epsilon_0(2b)^2}a_x$

$\large \vec{F_2}=\frac{-q^2}{4\pi\epsilon_0((2a)^2+(2b)^2)^{\frac{3}{2}}}(a_x2b+a_y2a)$

So now we have $\vec{F}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\Rightarrow\large \vec{F}=\frac{q^2}{16\pi\epsilon_0}[a_x(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})+a_y(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]$

$\Rightarrow|\vec{F}|=\frac{q^2}{16\pi\epsilon_0}\sqrt{[(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})]^2+[(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]^2}$

Now putting the known values:

$a = 10 \, \text {cm}$

$b = 18 \, \text {cm }$

$q = 1 \, \mathrm {\mu C}$

$\varepsilon_0 \approx 8.854 \cdot 10^{-12} \,\text{F}/\text{m}$ ,

We get $F=0.20028540143\approx\boxed{0.2}$