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I ran into trouble while reading example 3.8 from Griffiths' Introduction to Electrodynamics (4th ed.). The example is as follows: "An uncharged metal sphere of radius $R$ is placed in an otherwise uniform electric field $\mathbf{E}=E_0\mathbf{\hat{z}}$. Find the potential in the region outside the sphere".

Now, since the sphere is a conductor, we know it is an equipotential surface, so we can set it to zero. We can also deduce that, due to the external applied field, positive charge accumulates on the northern hemisphere of the sphere, while negative charge accumulates on the southern hemisphere. Due to these two facts, then, the entire $xy$ plane is at potential zero.

As for boundary conditions, we have:

$(i)\ V=0$ $\ \ \ \ \ \ \ \ \\ \ \ $ when $\ \ \ $ $r=R$

$(ii)\ V\rightarrow -E_0r\cos(\theta)$ $\ \ \ \ \ \ \ \ \ \ $ for $\ \ \ $ $r\gg R$

Now, the general solution for this problem is given by:

$$ V(r,\theta)=\sum_{l=0}^{\infty} \left(A_{l}*r^l+\frac{B_l}{r^{l+1}}\right)P_l(\cos(\theta)) $$

The next step is where something seems off to me. According to Griffiths, applying boundary condition $(i)$ yields:

$$ A_lR^l+\frac{B_l}{R^{l+1}}=0 $$

But why is this? Shouldn't it be

$$ \sum_{l=0}^{\infty} \left(A_{l}*R^l+\frac{B_l}{R^{l+1}}\right)P_l(\cos(\theta))=0 $$

instead?

Whys is Griffiths assuming that every term in the series is zero? Woudn't it be possible to have some terms be positive and others negative, so that the series as a whole converges to zero?

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Yes, but your last equation is valid for all $\theta$. You can use this fact and the orthogonality of the Legendre polynomials to conclude that all the coefficients are zero.

Indeed, for “any” function $f(z\in(-1,1))$, you can express it as: $$ f(z)=\sum_{l=0}^\infty \frac{2l+1}{2}f_lP_l(z)\\ f_l=\int_{-1}^1f(z)P_l(z)dz $$ Thus, if $f=0$, then $f_l=0$ for all $l\geq 0$. The converse is trivially true btw, so it’s actually and equivalence. Check out the completeness of the Legendre polynomials.

Hope this helps.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – driver99
    Nov 27, 2023 at 22:49

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