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The picture is a question from the book Intro to Electrodynamics by Griffiths. In question as you can see we want to find potential due to an infinite strip maintained at constant potential in the given region (slot).

My question is that the fourth boundary condition they used is ($V\to0$ as $x\to\infty$) which I think is wrong . Because when we deal with infinite charge distribution (like infinite line of charge or infinite sheet of charge) the convention that potential zero at infinity fails. For infinite charge distribution potential at infinity blows up. And hence we cannot use ($V\to0$ as $x\to\infty$) condition. So what is wrong with my argument here?

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  • $\begingroup$ No problem. What is it about Griffiths argument under the list of BCs that confuses you, though? If the potential is only non-zero at $x=0$, why shouldn't the potential drop off at $x\to\infty$? $\endgroup$ – Kyle Kanos Jan 11 '18 at 14:58
  • $\begingroup$ The convension of Setting zero potential at infinity fails in case of infinite charge distribution. This is line from the book "the symptom of trouble,in such cases is that potential blows up" $\endgroup$ – Vicky Shrimali Jan 11 '18 at 15:09
  • $\begingroup$ As the case of infinite strip at x=0 maintained at a specific potential is no different from infinite line of charge. $\endgroup$ – Vicky Shrimali Jan 11 '18 at 15:14
  • $\begingroup$ physics.stackexchange.com/q/311621/25301 $\endgroup$ – Kyle Kanos Jan 11 '18 at 15:18
  • $\begingroup$ Unfortunately that doesn't answer my question... $\endgroup$ – Vicky Shrimali Jan 11 '18 at 15:53
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The Answer is in Gauss's law. The Electric field for sheet is given by $$E = \frac{\sigma}{\epsilon_0}$$ Here $\sigma = Q/A$, where $Q=$ charge and $A=$ area of the sheet. So, In this case charge $Q$ and area $A$ both tends to infinity. Which results into finite value Electric field $E$. And, as $E$ is proportional to potential $V$, potential is also finite.

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    $\begingroup$ Potential at infinity is not finite in case of infinite charge distribution. Even if E is finite. $\endgroup$ – Vicky Shrimali Jan 11 '18 at 15:01
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The potential $V$ between the parallel planes at potential zero decreases exponentially to zero for $x\to\infty$ due to the shielding of the parallel metal plates at $V=0$. In this case the requirement$V\to0$ as $x\to\infty$ is actually not necessary. It ensures only that there is no infinite charge at infinity which would produce an potential solution which increases exponentially with $x$. In z-direction the potential is constant.

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  • $\begingroup$ Can you tell me how this case is different from infinite line charge in which potential doesn't drop to zero at infinity. $\endgroup$ – Vicky Shrimali Jan 12 '18 at 8:02
  • $\begingroup$ Vicky Shrimali - The difference arises due to the two metal plates, already at potential zero, extending to infinity. $\endgroup$ – freecharly Jan 12 '18 at 14:18
  • $\begingroup$ So, you mean the two plates is the reason of zero potential at infinity. If you're 100 % sure then thanks man. It makes sense. $\endgroup$ – Vicky Shrimali Jan 12 '18 at 15:32
  • $\begingroup$ @Vicky Shrimali - I am completely sure! $\endgroup$ – freecharly Jan 12 '18 at 15:41

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