# Where should I set the zero of electric potential in presence of an external electric field?

I put a charged or uncharged conducting sphere in an initially uniform electric field. I know that the potential is a constant inside and over the surface of the sphere, and approximately the same as the potential due to the external field far from the sphere. But then when should I set the zero of potential if it's not at infinity?

• They did tell you that the position of the zero of the electrostatic potential is ultimately irrelevant, and that only potential differences are physical, right? – Emilio Pisanty Nov 8 '16 at 15:12

• You set the potential to zero at one point (which in the end is just for convenience since, as Emilio Pesanty mentions, only potential differences matter, but by having a single point of reference you can give each point a number which saves you from taking differences again and again). If you don't have your sphere there will be no potential difference across the xy-plane, so the whole plane will have potential zero. If you do have your sphere, the potential will change as you move closer to the sphere: in R. Physics' answer, set $\theta=\pi$, i.e. $\cos \theta = 0$ and you get a $1/r$ term. – Lukas Berns Nov 15 '16 at 0:30
• If however the sphere is uncharged ($Q=0$) the whole x-y plane has the same potential. – Lukas Berns Nov 15 '16 at 0:32
Due to symmetry, the general solution for the Laplace equation is $$V\left(\theta,r\right)=\sum_{l=0}^{\infty}P_{l}\left(\cos\theta\right)\left(A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}\right)$$ the restriction for our system is $$\frac{\partial V}{\partial z}|_{r\rightarrow\infty}=-E$$ this implies $$V\left(\theta,r\right)_{\lim r\rightarrow\infty}=V_{\infty}-Er\cos\theta$$ so the solution is then $$V\left(\theta,r\right)=V_{\infty}-Er\cos\theta+\sum_{l=0}^{\infty}P_{l}\left(\cos\theta\right)\left(\frac{B_{l}}{r^{l+1}}\right)$$ Now we have another restriction, this is that the potential on the sphere must be constant $$V\left(\theta,R\right)=V_{S}$$ so $$V_{S}=V_{\infty}-ER\cos\theta+\sum_{l=0}^{\infty}P_{l}\left(\cos\theta\right)\left(\frac{B_{l}}{r^{l+1}}\right)$$ the only way how this can be done is if we have $B_{l}=0$ for $l>1$ then if the sphere has radius $R$ $$V_{S}=V_{\infty}-ER\cos\theta+\frac{B_{0}}{R}+\frac{B_{1}}{R^{2}}\cos\theta$$ this implies( because the potential is constant at the sphere) $$ER^{3}=B_{1}$$ and also $$\left(V_{S}-V_{\infty}\right)R=B_{0}$$ so we have $$V\left(\theta,r\right)=V_{\infty}-Er\cos\theta+\frac{\left(V_{S}-V_{\infty}\right)R}{r}+\frac{ER^{3}}{r^{2}}\cos\theta$$ we can relate the term proportional to $\frac{1}{r}$ to the charge pf the sphere: $$V_{S}-V_{\infty}=\frac{Q}{4\pi\epsilon_{0}R}$$ this is the important point, you can fix the zero potential at any point, but the important fact is that the difference between the potential at infinity and the potential at the sphere is proportional to the charge in the sphere. The constant $V_{\infty}$ is just the constant $A_{0}$ in the general solution. I call it $V_{\infty}$ because it is a constant term that survives at infinity for any value of the electric field. The potential in the sphere is $V_{S}$. There is a formal way to calculate the net charge of the field by using the Maxwell equation $\int\mathbf{E}\cdot\hat{n}dA=\frac{q}{\epsilon_{0}}\rightarrow E_{\perp}^{+}-E_{\perp}^{-}=\frac{\sigma}{\epsilon_{0}}$. If you calculate the gradient of the potential to get the normal component of the electric field and then from there you get the charge density you will be able to get the total charge of the sphere after an integration over the surface. This is the correct way to do it, if you do that calculation you will get that $$\left(V_{S}-V_{\infty}\right)4\pi\epsilon_{0}R=Q$$ another way to see this is if you start to reduce the value of your electric field, then the only terms that survive is $$V\left(\theta,r\right)=V_{\infty}+\frac{\left(V_{S}-V_{\infty}\right)R}{r}$$ thus we have a potential that decreases as $\frac{1}{r}$ ,this is just the potential of a charged particle and we know by Gauss theorem that it is equal to $$V\left(\theta,r\right)=V_{\infty}+\frac{Q}{4\pi\epsilon_{0}r}$$ where $V_{\infty}$ is just the potential at inifinity, from here we get $$\left(V_{S}-V_{\infty}\right)4\pi\epsilon_{0}R=Q$$ perhaps you can say that there could be a change of the total charge of the sphere induced by the term $\frac{ER^{3}}{r^{2}}\cos\theta$ when the field is turned on, but check out the form of this term, it decreases as $\frac{1}{r^{2}}$ just like a dipole and we also have the $\cos\theta$ which is caracteristic of the dot product, so this term is only produced from opposite charges to the opposite sides of the sphere, so the nete contribution to the charge is zero.