# Method of image charges [closed]

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.

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

• I'm voting to close this question as off-topic because check-my-work questions are off-topic. Jun 13, 2015 at 11:18
• @ACuriousMind That's wonderful. All my effort goes down the drain. (I didn't mean necessarily a math error - perhaps my physics is rusty and something is wrong in my conclusions etc.?). BTW I presented a method of solving a problem - my question in itself has a value to the readers, for being able to understand physics better. Which can be enhanced with a good answer...
– Dor
Jun 13, 2015 at 11:32
• Please note that this is not a homework help site. See How do I ask homework questions on Physics Stack Exchange? and Should any check my work questions be made on topic? posts on meta for more information. Jun 13, 2015 at 15:41
• Jun 17, 2015 at 15:50
• Dor, as I understand it you are confused because you think the expression for the total charge should be finite and convergent. Right? Now, that doesn't necessarily mean the method of image charges doesn't apply to this setup. I think it would make your question better if you explain in more detail why you think the divergence of $\sqrt{r^2 + b^2}$ implies that the method of image charges isn't valid here. Jun 18, 2015 at 10:40

• You wrote "first-term in your dipole field is wrong". Can you please elaborate? I took only the z component, therefore $r^2$ is cancelled out..