[Physics] Method of image charges

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In an attempt to understand the method of image charges, I'll try to calculate the total charge on a grounded conducting plane – with electric dipole & point charge.

Given:

1. Point charge $Q$, located at $(0,0,a)$.
2. Electric dipole $\vec p = (0,0,p)$, located at $(0,0,b)$.
3. Grounded & infinite conducting plane, at $z=0$ (parallel to XY plane)

My work

Employing the method of image charges:

Obviously, by logic, it is the total charge of the dipole + the charge $Q$, which amount to merely $Q$ (because the dipole has two charges with equal magnitude and opposite sign.)

I tried to find the answer, $Q$, mathematically, by applying Gauss law on the plane, having $\vec E \ne 0$ only on the positive side of $z$.

So first I tried to find expressions for the $z$ component of the electric field at a point which is very close to the grounded plane:

Electric field of a Dipole: $$\vec E (\,\vec r\,) = \frac{
3(\,\vec p \cdot \hat r \,)\hat r – \vec p
}{
4\pi \epsilon_0 \cdot \lvert \vec{ r\,} \rvert^3
}$$

While $\vec r = (x,y,0) – (0,0,b) = (x,y,-b)$, its $z$ component is:
$$
\vec E_{z\,, \,dipole} = \frac{1}{4\pi \epsilon_0} \cdot \left( \frac{3pb^2}{\sqrt{x^2+y^2+b^2}} – \frac{p}{(x^2+y^2+b^2)^{3/2}} \right)
$$

Electric field of the point charge (using Coulomb's law):

$$E_Q = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{ (x^2+y^2+a^2)^{3/2}} \cdot (x,y,-a) $$

So by gauss law, the total charge on the plane is:

$$
\begin{align}
Q_{in} &= \epsilon_0 \oint_S \vec E d \vec S = \\
&= \epsilon_0 \frac{1}{4\pi \epsilon_0} \int\limits_{0}^{2\pi} \int\limits_{0}^{\infty} \left( \frac{3pb^2}{\sqrt{r^2 + b^2}} – \frac{p}{(r^2 + b^2)^{3/2}} – \frac{a Q}{(r^2+a^2)^{3/2}} \right) r \, dr \, d\theta = \\
&= \frac{1}{4\pi} \cdot 2\pi \left( 3pb^2 \cdot \sqrt{r^2 + b^2} + \frac{p}{\sqrt{r^2 + b^2}} + \frac{a Q}{\sqrt{r^2 + a^2}} \right) \bigg{|}_{r=0}^{\infty} \\
\end{align}
$$
But as you can see, the first expression $\left( 3pb^2 \cdot \sqrt{r^2 + b^2} \right)$ diverges.

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Please explain why the method of image charges makes sense in this scenario?

Answer 1

There is a much simpler way to find the charge on the conductor. Using the method of images, the image charges are just the negative of the charges above the plane. Therefore, the total charge on the surface of the plane is -Q. The first-term in your dipole field is wrong. The unit vectors have unit value and do not cancel r squared in the denominator.