1
$\begingroup$

Considering the case where we have a grounded sphere (so the potential on the surface is 0) and the following charge system:Grounded sphere - image charge problem

Now, I'm trying to solve the case for when, instead of a grounded sphere, we have a charged one, which I assumed to be the same as before, but with an extra charge Q at the centre of the sphere. Solving the Laplace equation for $R\leq r$, I got that: $$\phi (x,y,z)=\frac{1}{4\pi \epsilon_0}[\frac{q}{\sqrt{(D-x)^2 + y^2+z^2}}+\frac{qR}{D\sqrt{(x-b)^2 + y^2+z^2}}+\frac{Q}{\sqrt{x^2 + y^2+z^2}}]$$

But now how do I find the charge distribuition around the surface of sphere? I was thinking of changing everything to spherical coordinates and then solving the equation which comes from Gauss' Theorem: $\sigma =\epsilon_0 \frac{\delta \phi}{\delta r}$ at r=R and: $$\frac{\delta \phi}{\delta r}=\frac{\delta \phi}{\delta x}\frac{\delta x}{\delta r}+\frac{\delta \phi}{\delta y}\frac{\delta y}{\delta r}+\frac{\delta \phi}{\delta z}\frac{\delta z}{\delta r}$$

But isn't there another way of doing this?

$\endgroup$
2
  • 1
    $\begingroup$ wikipedia talks about this case I didn't went trough it but it seems it could serve as a hint. $\endgroup$
    – Dabed
    Mar 9 '20 at 19:46
  • $\begingroup$ they just give the result, but it doesn't say how they derived it $\endgroup$
    – Rye
    Mar 10 '20 at 9:55
2
$\begingroup$

The system can be written as

$\phi (r,\theta) =\frac{1}{4\pi \epsilon_0}[\frac{Q}{|r-d|}+\frac{Q_i}{|r-d_i|}] $

Now for $r=a$ we want

$ 0=\phi (r=a,\theta) =\frac{1}{4\pi \epsilon_0}[\frac{Q_i}{|a-d_i|}+\frac{Q}{|a-d|}]\\ \Rightarrow Q^2|a-d_i|^2=Q_i^2|a-d|^2\\ \Leftrightarrow Q^2(a^2+d_i^2-2ad_i\cos(\theta))=Q_i^2(a^2+d^2-2ad\cos(\theta))\\ \Leftrightarrow Q^2(a^2+d_i^2)-Q_i^2(a^2+d^2)=2a\cos(\theta)(Q^2d_i-Q_i^2d) $

We would like the induced charge to be homogeneous on the sphere $\phi (r=a,\theta)=\phi (r=a)$ so we need that

$Q^2d_i-Q_i^2d=0\quad (1)\Rightarrow Q^2(a^2+d_i^2)-Q_i^2(a^2+d^2)=0\quad (2)$

therefore

$0=_{2}(a^2+d_i^2)-\frac{Q_i^2}{Q^2}(a^2+d^2)\\ =_{1}(a^2+d_i^2)-\frac{d_i}{d}(a^2+d^2)\\ =d_i^2-\frac{(a^2+d^2)}{d}d+a^2\\ =(d_i-\frac{a^2}{d})(d_i-d) $

The solution $d_i=d$ is outside the sphere right over the other charge so instead we take $d_i=\frac{a^2}{d}$ which implies $Q_i=\pm\frac{a}{d}Q$ but both charges can't be positive so we take $Q_i=-\frac{a}{d}Q$

Notice that this means we have $d_id=a^2$ which is called also an sphere inversion $OP\times OP^{\prime} =r^2$ (if we where working in $\mathbb{C}$ instead of $\mathbb{R^2}$ this is would read as $z\overline{z}=\|z\|$) and further can be read at the wikipedia article for the image method

Putting this back in our system we get

$\phi (r,\theta)\\ =\frac{1}{4\pi \epsilon_0}[\frac{Q}{|r-d|}+\frac{Q_i}{|r-d_i|}]\\ =\frac{Q}{4\pi \epsilon_0}[\frac{1}{|r-d|}+\frac{-a/d}{|r-a^2/d|}]\\ =\frac{Q}{4\pi \epsilon_0}[\frac{1}{|r-d|}+\frac{-1}{|rd/a-a|}]\\ =\frac{Q}{4\pi \epsilon_0}[\frac{1}{\sqrt{r^2+d^2-2rd\cos(\theta)}}+\frac{-1}{\sqrt{r^2d^2/a^2+a^2-2rd\cos(\theta)}}]\\ $

The derivative evaluated at $r=a$ gives

$\phi_r (r,\theta)|_{r=a}\\ =\frac{Q}{4\pi \epsilon_0}[-\frac{d\cos(\theta)-r}{(r^2+d^2-2rd\cos(\theta))^{3/2}}+\frac{d\cos(\theta)-rd^2/a^2}{r^2d^2/a^2+a^2-2rd\cos(\theta))^{3/2}}]|_{r=a}\\ =\frac{Q}{4\pi \epsilon_0}[-\frac{d\cos(\theta)-a}{(a^2+d^2-2ad\cos(\theta))^{3/2}}+\frac{d\cos(\theta)-d^2/a}{(d^2+a^2-2ad\cos(\theta))^{3/2}}]\\ =\frac{Q}{4\pi \epsilon_0}[\frac{a^2-d^2}{a(d^2+a^2-2ad\cos(\theta))^{3/2}}]\\ $

Finally we want the total charge on the surface:

$Q_{\text{total}}=\int\sigma(\theta)d\Omega\\ =\int_{-\pi}^{\pi}\int_0^{\pi}\epsilon_0\phi_r(r=a,\theta)a^2\sin(\theta)d\theta d\varphi\\ =\int_{-\pi}^{\pi}\int_{0}^{\pi}\epsilon_0(\frac{Q}{4\pi \epsilon_0}[\frac{a^2-d^2}{a(d^2+a^2-2ad\cos(\theta))^{3/2}}])a^2\sin(\theta)d\theta d\varphi\\ =\frac{a(a^2-d^2)Q}{4\pi}\int_{-\pi}^{\pi}d\varphi\int_{-\pi}^{\pi}[\frac{\sin(\theta)}{a(d^2+a^2-2ad\cos(\theta))^{3/2}}]d\theta\\ =\frac{a(a^2-d^2)Q}{2}\int_{0}^{\pi}[\frac{\sin(\theta)}{(d^2+a^2-2ad\cos(\theta))^{3/2}}]d\theta\\ =\frac{a(a^2-d^2)Q}{2}[\frac{1/ad}{\sqrt{d^2+a^2-2ad\cos(\theta)}}]|_{0}^{\pi}\\ =\frac{(a^2-d^2)Q}{2d}[\frac{1}{\sqrt{d^2+a^2-2ad}}-\frac{1}{\sqrt{d^2+a^2+2ad}}]\\ =\frac{(a^2-d^2)Q}{2d}[\frac{1}{\sqrt{(d-a)^2}}-\frac{1}{\sqrt{(d+a)^2}}]\\ =\frac{(a^2-d^2)Q}{2d}[\frac{1}{(d-a)}-\frac{1}{(d+a)}]\\ =\frac{(a^2-d^2)Q}{2d}[\frac{2a}{(d^2-a^2)}]\\ =\frac{-aQ}{d}\\ $

Where the factor $\frac{-a}{d}$ represents the fraction of charge induced by the particle at distance $d$ over the sphere of radius $a$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.