Since this problem consists of multiple parts and one needs to see all of them to understand the problem i'm going to list out all of them:
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Consider a uniformly charged spherical shell of radius $R$ with surface charge density $\sigma_0$. Compute the potential outtside the shell, $r>R$.
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Now consider we slighttly squzee the sphere (without changing the surface charge density) along one axis so that the surface is now located at
$$r=\frac{R}{\left(b^2\cos^2(\beta)+\sin^2(\beta)\right)^{1/2}}$$
where $b$ is a number such that $0<b^2-1 \ll 1$ and $\beta$ is the angle between a point on the surface and the symmetry axis of the system. Find the potential for $r<R$ neglecting terms $O(\epsilon^2)$ where $\epsilon=b^2-1$.
Hint: Recall the potential in terms of the charge volume density $\rho$ is
$$\Phi(\vec{r})=\frac{1}{4\pi\epsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\int \mathrm d^3r'(r')^n\rho (\vec{r}')P_n(\cos\alpha)$$
where $\alpha$ is the angle between $\vec{r}$ and $\vec{r}'$.
The first question is simple enough, but I don't really understand the second question at all. Appreciate any help!
EDIT: I shall provide the (my) solution to the first part of the question:
The potential of a uniformly charged spherical shell of radius $R$ with surface charge density $\sigma_0$ outside of $R$ is simply $V=\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$, for $r > R$.
To formulate my question a bit better:
Since the potential in electrostatics is generally given by the equation:
$$V(r)=\frac{1}{4\pi\epsilon_0}\int \frac{dq}{r}$$
or in this case (using spherical coordinates):
$$V(r)=\frac{\sigma_0}{4\pi\epsilon_0}\int \frac{dA}{r}=\frac{\sigma_0}{4\pi\epsilon_0}\int \frac{1}{r}R^2d\theta d\phi$$
Now if the sphere is "slightly squeezed", I can no longer use the above equation, and the shape of the sphere becomes somewhat of two (if I can assume again) U-shaped magnets with their open ends closed together. But then i don't understand the part where it says "the surface is now located at
$$r=\frac{R}{\left(b^2\cos^2(\beta)+\sin^2(\beta)\right)^{1/2}}$$"
Is this $r$ referring to the center of the object?
Best Answer
The potential at the field point is therefore $$ \Phi(r,\theta)=\frac{\sigma_{0}}{4\pi\epsilon_{0}}\int_{D}\frac{{\rm d}D}{|\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}|}=\frac{\sigma_{0}}{4\pi\epsilon_{0}} \int_{0}^{2\pi}{\rm d}\phi\int_{0}^{\pi}{\rm d}\beta\,\,\frac{r_{\beta}^2\,\sin\beta}{|\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}|} \,, \tag{2} $$ where $D$ denotes the surface of the deformed sphere. To shorten the expressions, we now define $$ \zeta=\frac{\epsilon}{2}\cos^2\beta\,,\text{ so that } \zeta=O(\epsilon)\,. \tag{3} $$ From the left hand figure, the angle between the position vectors $\mathbf{r_\beta}$ and $\mathbf{r}$ is $\gamma,$ and from the cosine rule we can write $$ |\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}|=\sqrt{r_{\beta}^2+r^2-2rr_{\beta}\cos\gamma}\,; \tag{4} $$ and we now need to find an expression for $\cos\gamma.$ The Cartesian coordinates of the position vectors in the figure are $$ \mathbf{r_{\beta}}=r_{\beta}\left(\sin\beta\cos\phi,\sin\beta\sin\phi,\cos\beta\right)\quad\text{and}\quad \mathbf{r}=r\left(\sin\theta\cos\mu,\sin\theta\sin\mu,\cos\theta\right)\,. \tag{5} $$ By the azimuthal symmetry of the problem, the potential is independent of the angle $\phi,$ and we can rotate the $x$-$y$ axes about the vertical $z$ axis so that $\mu\,$ becomes zero. With $\mu=0$ we then have $\cos\mu=1$ and $\sin\mu=0,$ and the field point vector $$\mathbf{r}=r(\sin\theta,0,\cos\theta) \tag{6} $$ is now chosen to lie in the $x$-$z$ plane. The dot product of the position vectors is then $$ \mathbf{r_{\beta}}\cdot\mathbf{r}=rr_{\beta}(\sin\theta\sin\beta\cos\phi+\cos\theta\cos\beta)=rr_{\beta}\cos\gamma\,, \tag{7} $$ and therefore the required expression for $\cos\gamma$ is $$ \cos\gamma=\sin\theta\sin\beta\cos\phi+\cos\theta\cos\beta\,. \tag{8} $$
From $(1)$ and $(3),$ we have on expanding $r_{\beta}^2$ using the binomial theorem, $$ r_{\beta}=R(1-\zeta)+O(\epsilon^2)\,, \ \ \ r_{\beta}^2=R^2(1-2\zeta)+O(\epsilon^2)\,, \tag{9} $$ and from $(4)$ and $(9)$ we can write $$ |\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}| =\sqrt{R^2(1-2\zeta)+r^2-2rR(1-\zeta)\cos\gamma} \,, $$ or, on grouping the $\zeta$ terms, $$ |\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}| =\sqrt{R^2+r^2-2rR\cos\gamma-2\zeta R(R-r\cos\gamma)}\,. \tag{10} $$ Now put $$ \mathfrak{R}=\sqrt{R^2+r^2-2rR\cos\gamma}\quad\text{and}\quad \lambda=\sqrt{2\zeta R(R-r\cos\gamma)}\,, $$ and note that $\lambda^2=O(\epsilon)$ because $\zeta=O(\epsilon)$ from $(3)$. We can then write $$ |\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}|= \sqrt{\mathfrak{R}^{2}-\lambda^{2}}=\mathfrak{R}\,\sqrt{1-\left(\lambda/\mathfrak{R}\right)^{2}}\,, \tag{11} $$ and by binomially expanding the square root, $$ \frac{1}{|\mathbf{r_{\beta}}-\mathbf{r} \mspace{1mu}|} =\frac{1}{\mathfrak{R}}\left(1-\left(\lambda/\mathfrak{R}\right)^{2}\,\right)^{\!-\frac{1}{2}}=\frac{1}{\mathfrak{R}}\left(1+\tfrac{1}{2}\left(\lambda/\mathfrak{R}\right)^{2}\,\right)+O(\epsilon^2)\,. \tag{12} $$ Denoting the integrand of $\Phi(r,\theta)$ by $\mathcal{I},$ and using $(12)$, its value is now $$\mathcal{I}=\frac{r_{\beta}^2\,\sin\beta}{|\mathbf{r_{\beta}}-\mathbf{r\mspace{1mu}|}}=\frac{r_{\beta}^2\sin\beta}{\mathfrak{R^{\vphantom{Y}}}} \left(\!1+\tfrac{1}{2}\left(\lambda/\mathfrak{R}\right)^{2}\,\right)+O(\epsilon^2)\,. \tag{13} $$ The reciprocal radius $\mathfrak{R}^{-1}$ may be expanded into a series of Legendre polynomials to give $$ \frac{1}{\mathfrak{R}}=\frac{1}{\sqrt{R^2+r^2-2rR\cos\gamma}}= \frac{1}{R}\sum_{n=0}^{\infty}\left(\frac{r}{R}\right)^{\!n}P_{n}(\cos\gamma)\,, \tag{14} $$ and keeping terms up to $(r/R)^2$, we write the reciprocal radius as $$ \frac{1}{\mathfrak{R}}=\frac{S}{R}\,,\tag{15} $$ where $S$ denotes the polynomial sum $$ S =P_{0}(\cos\gamma)+\left(\frac{r}{R}\right)P_{1}(\cos\gamma)+\left(\frac{r}{R}\right)^{\!2 }P_{2}(\cos\gamma)\,. \tag{16} $$ For a sphere the potential inside is constant for a uniform surface charge, and since $\epsilon\ll 1$ for the deformed sphere, the deviation from the constant value will be small. The variations of the potential with $r$ will also be small, and we therefore neglect powers of $r/R\,$ higher than the quadratic terms.
Taking $P_{0}(x)=1$, $P_{1}(x)=x$, $P_{2}(x)=\frac{1}{2}\left(3x^2-1\right)$, we get $$ S =1+\frac{r\cos\gamma}{R}+\left(\frac{r}{R}\right)^{\!2}\frac{1}{2}\left(3\cos^2\gamma-1\right)\,. \tag{17} $$ Equations $(9)$ and $(13)$, and the preceding definition of $S$, then allow the integrand to be written as $$ \mathcal{I}=R^2(1-2\zeta)\sin\beta\left\{ \frac{S}{R}\left(1+\frac{1}{2}\frac{\lambda^2 S^2}{R^2}\right)\!\right\} +O(\epsilon^2)\,. \tag{18} $$ The potential is then $$ \Phi(r,\theta)=\frac{\sigma_{0}}{4\pi\epsilon_{0}} \int_{0}^{2\pi}{\rm d}\phi\int_{0}^{\pi}{\rm d}\beta\,R^2(1-2\zeta)\,\sin\beta\left\{ \frac{S}{R}\left(1+\frac{1}{2}\frac{\lambda^2 S^2}{R^2}\right)\!\right\} \,, \tag{19} $$ where $$ \cos\gamma=\sin\theta\sin\beta\cos\phi+\cos\theta\cos\beta\,, \tag{20} $$ $$ \zeta=\frac{1}{2}\cos^2\beta\,, \tag{21} $$ $$\lambda^2=2\zeta R(R-r\cos\gamma)\,,\tag{22}$$ and $S$ is the Legendre expansion in $(17)$. The integrations over $\phi$ and $\beta$ are easy to calculate, since the expanded integrand now contains only powers of trigonometric functions, and with a symbolic algebra system it is easy to obtain higher order terms. We get $$ \Phi(r,\theta)=\frac{\sigma_{0}}{4\pi\epsilon_{0}}\left\{4\pi R-\frac{2\pi\epsilon R}{3} +\frac{\pi\epsilon r^2}{15R}\left(\vphantom{Y^{Y^Y}}1+3\cos 2\theta\,\right)\right\} \tag{23} \,, $$ and on cancelling the $4\pi$'s, $$ \Phi(r,\theta)=\frac{\sigma_{0}R}{\epsilon_{0}}\left\{1-\frac{\epsilon}{6}+\frac{\epsilon}{60}\left(\frac{r}{R}\right)^{\!2}\left(\vphantom{Y^{Y^Y}}1+3\cos 2\theta \,\right)\right\} \tag{24} \,, $$ or, in terms of the total charge $Q=4\pi R^2\sigma_{0}$ on the sphere, $$ \Phi(r,\theta)=\frac{Q}{4\pi\epsilon_{0}R}\left\{1-\frac{\epsilon}{6}+\frac{\epsilon}{60}\left(\frac{r}{R}\right)^{\!2}\left(\vphantom{Y^{Y^Y}}1+3\cos 2\theta\, \right)\right\} \tag{25} \,. $$ The integrand could have been expressed in terms of spherical harmonics and use made of the addition theorem for $P_{n}(\cos\gamma)$ and the orthogonal properties of the spherical harmonics, but the low order trigonometric functions in this problem are easily integrated without their use, since only powers of trigonometric functions appear in the integrand.
Edit 1: I have added in the comment here.
For a short answer, without the $θ$ variation, the potential inside the sphere is almost constant, so take the field point $\mathbf{r_{\beta}}$ at the centre of the sphere. Then the denominator becomes $r_{\beta}$ and the integrand becomes $r^{2}_{β} \sin\beta/r_{\beta},$ and taking $r_{\beta}$ from $(1)$ and integrating, we get the $1−\epsilon/6$ term immediately.
Edit 2: I have expanded the integrand from (18) here.
Without using spherical harmonics, the value of $\mathcal{I}$ in $(18)$ in terms of trigonometric powers is $$\begin{align} \mathcal{I}&=-{{\cos ^4\beta\,\sin \beta\,\epsilon^2\,R}\over{2}}\\ &- {{\cos ^2\beta\,\sin\beta\,\epsilon\,R}\over{2}}\\ &+\sin \beta\,R \\ &-{{9\,\cos ^4\beta\,\sin ^3\beta \,\epsilon^2\,\cos ^2\phi\,r^2\,\sin ^2\theta}\over{4\,R}}\\ &+{{3\,\cos^2\beta\,\sin ^3\beta\,\epsilon\,\cos ^2\phi\,r^2\,\sin ^2\theta }\over{4\,R}} \\ &+{{3\,\sin ^3\beta\,\cos ^2\phi\,r^2\,\sin ^2 \theta}\over{2\,R}} \\ &-{{9\,\cos ^5\beta\,\sin ^2\beta\,\epsilon^2\,\cos \phi\,r^2\,\cos \theta\,\sin \theta}\over{2\,R}} \\ &+{{3\,\cos^3\beta\,\sin ^2\beta\,\epsilon\,\cos \phi\,r^2\,\cos \theta\,\sin\theta}\over{2\,R}} \\ &+{{3\,\cos \beta\,\sin ^2\beta\,\cos \phi\, r^2\,\cos \theta\,\sin \theta}\over{R}}\\ &-{{9\,\cos ^6\beta\, \sin \beta\,\epsilon^2\,r^2\,\cos ^2\theta}\over{4\,R}}\\ &+{{3\,\cos ^4 \beta\,\sin \beta\,\epsilon\,r^2\,\cos ^2\theta}\over{4\,R}}\\ &+{{3\,\cos ^ 2\beta\,\sin \beta\,r^2\,\cos ^2\theta}\over{2\,R}}\\ &+{{3\,\cos ^4 \beta\,\sin \beta\,\epsilon^2\,r^2}\over{4\,R}}\\ &-{{\cos ^2\beta\,\sin \beta \,\epsilon\,r^2}\over{4\,R}}\\ &-{{\sin \beta\,r^2}\over{2\,R}}\\ &-{{5\,\cos ^4 \beta\,\sin ^4\beta\,\epsilon^2\,\cos ^3\phi\,r^3\,\sin ^3\theta }\over{4\,R^2}}\\ &+{{5\,\cos ^2\beta\,\sin ^4\beta\,\epsilon\,\cos ^3\phi\, r^3\,\sin ^3\theta}\over{4\,R^2}}\\ &-{{15\,\cos ^5\beta\,\sin ^3 \beta\,\epsilon^2\,\cos ^2\phi\,r^3\,\cos \theta\,\sin ^2\theta }\over{4\,R^2}}\\ &+{{15\,\cos ^3\beta\,\sin ^3\beta\,\epsilon\,\cos ^2\phi \,r^3\,\cos \theta\,\sin ^2\theta}\over{4\,R^2}}\\ &-{{15\,\cos ^6 \beta\,\sin ^2\beta\,\epsilon^2\,\cos \phi\,r^3\,\cos ^2\theta\,\sin \theta}\over{4\,R^2}}\\ &+{{15\,\cos ^4\beta\,\sin ^2\beta\,\epsilon\,\cos \phi\,r^3\,\cos ^2\theta\,\sin \theta}\over{4\,R^2}}\\ &+{{3\, \cos ^4\beta\,\sin ^2\beta\,\epsilon^2\,\cos \phi\,r^3\,\sin \theta }\over{4\,R^2}}\\ &-{{3\,\cos ^2\beta\,\sin ^2\beta\,\epsilon\,\cos \phi\,r^ 3\,\sin \theta}\over{4\,R^2}}\\ &-{{5\,\cos ^7\beta\,\sin \beta\,\epsilon^2 \,r^3\,\cos ^3\theta}\over{4\,R^2}}\\ &+{{5\,\cos ^5\beta\,\sin \beta \,\epsilon\,r^3\,\cos ^3\theta}\over{4\,R^2}}\\ &+{{3\,\cos ^5\beta\,\sin \beta\,\epsilon^2\,r^3\,\cos \theta}\over{4\,R^2}}\\ &-{{3\,\cos ^3\beta\, \sin \beta\,\epsilon\,r^3\,\cos \theta}\over{4\,R^2}}\,. \end{align} $$