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Electric potential at a point is defined as the amount of work done in bringing a test charge from infinity to that point..my question is that if we don't know where this point of infinity lies then how can we calculate the potential at a point in influence of electric field and if we do know where it lies can you tell me where.. Please also tell me the assumptions while calculating electric potential at a point Also how do we calculate electric potential at a point for practical purposes with respect to a common zero potential point

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  • $\begingroup$ What do you mean by "where this point of infinity lies"? Are you just asking for a more rigorous description of "infinitely far away"? $\endgroup$ – BioPhysicist May 28 at 14:34
  • $\begingroup$ Well..no..my question is that if we define potential with respect to infinity then how can we calculate it as we cannot certain the position of our reference point(in this case infinity) in the first place ie. How can we calculate the work done.. $\endgroup$ – user265825 May 28 at 14:43
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In general, the potential difference between two points $A$ and $B$ with position vectors $\mathbf a$ and $\mathbf b$ respctively is $$V(\mathbf b)-V(\mathbf a)=-\int_C\mathbf E\,\cdot\text d\mathbf x$$ where $C$ is any path in space from $A$ to $B$.

So now let's say we want to find the potential at $B$ relative to "infinity". When we talk about using "infinity" as a reference point, what we mean is that you first determine $V(\mathbf b)-V(\mathbf a)$, then you take the limit as $\mathbf a$ goes to infinity (any point at infinity will do in general, although certain counterexamples exist in specific cases). Furthermore, if we define "infinity" to be the zero point of potential, then $$\lim_{\mathbf a\to\infty}V(\mathbf a)\equiv0$$ and so we can just talk about the potential at B since $$\lim_{\mathbf a\to\infty}V(\mathbf b)-V(\mathbf a)=V(\mathbf b)$$

Whenever infinities are used, the more formal way to handle them is to take limits. This is usually understood though, so you will often just hear people say "at infnity". You don't need to know exactly where this point is. Any point very far away will do (unless there are some issues with the charge distribution at infinity, e.g. for an infinite line charge).

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  • $\begingroup$ I do not agree that $\infty$ is taking limit to all coordinates to $\infty$, because $(0, \infty ,0)$ is also at $\infty$. I always thought that the infinity was only defined when the field,(i.e. potential) depended on one variable ($r$,$z$). $\endgroup$ – MarcoCiafa May 28 at 15:02
  • $\begingroup$ @MarcoCiafa Good point. I will edit $\endgroup$ – BioPhysicist May 28 at 15:03
  • $\begingroup$ I'm not sure I agree with the "one variable" argument for infinity, though. the potential due to a dipole depends both on $r$ and $\theta$, and we can still define the potential at infinity. $\endgroup$ – Philip May 28 at 15:05
  • $\begingroup$ @Philip Is that to me or MarcoCiafa? $\endgroup$ – BioPhysicist May 28 at 15:19
  • $\begingroup$ @MarcoCiafa I agree with Phillip, I am unsure what you are talking about with the field depending on only one variable. $\endgroup$ – BioPhysicist May 28 at 16:03
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The electric potential is specified up to a constant. In order to remove any ambiguity, the electric potential is usually defined to be $$V(r) = -\int_{r_\text{ref}}^r \vec{E}.\text{d}\vec{l}.$$

From this definition, the potential at the point $r_\text{ref}$ is always "zero". For example, consider the case of a point charge at the origin. The Electric Field is given by $$\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\hat{r},$$ and so from our definition above, $$V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r} - \frac{1}{4\pi\epsilon_0} \frac{q}{r_\text{ref}}.$$

A simple choice of $r_\text{ref}$ would be $r_\text{ref}=\infty$, which removes the second term in the equation above, but really, any point could be used. This is also the case for most "finite" charge distributions, the field falls off sufficiently fast very far away, so that the value of the second term above at the point $r = \infty$ is zero, which is why for finite charge distributions we use $r_\text{ref} = \infty$, and $$V(r) = -\int_{\infty}^r \vec{E}.\text{d}\vec{l}.$$

For an infinite line of charge, there is some charge at infinity, and so there's no reason for the field to die down at infinity in every direction (certainly, along the infinite wire it doesn't). And therefore, a different reference point is needed. I believe (unless I'm wrong) that the reference point is usually taken to be $r=1$ in some units, since the Electric Field for an infinite line charge is (in cylindrical coordinates) $$\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r},$$ and so $$V(r) = \int_{r_\text{ref}}^r \frac{\lambda}{2 \pi \epsilon_0 r} \text{d}r = \frac{\lambda}{2 \pi \epsilon_0} \left(\log{r} - \log{r_\text{ref}}\right) = \frac{\lambda}{2 \pi \epsilon_0} \left(\log{r} - \log{1}\right) = \frac{\lambda}{2 \pi \epsilon_0} \log{r}.$$

But it doesn't matter where you choose your $r_\text{ref}$ (as long as the value of the integral there is finite!); the potential there is by definition zero.

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