# Finding the charge distribution on a charged sphere - image method

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

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?

• wikipedia talks about this case I didn't went trough it but it seems it could serve as a hint. Commented Mar 9, 2020 at 19:46
• they just give the result, but it doesn't say how they derived it
– Rye
Commented Mar 10, 2020 at 9:55

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 $$a$$