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I just began studying electrostatics in university, and I didn't understand completely why the electric potential due to a conducting sphere is

$$ V(\vec{r})=\begin{cases} \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}, & \text{if $r \le R$}.\\ \\ \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}, & \text{if $r \gt R$}. \end{cases} $$

Where Q is the total charge and R is the radius of the sphere (the sphere is located at the origin).

I only understand the second part of this equation (when $r > R$).

I know Gauss Law. know the charges go to the surface. I know the electric field strictly inside it must be zero. (I also know the electric field is not defined for a point that lies exactly in the surface). And I know $\vec{E} = -\nabla{V}$.

Therefore, I know the electric potiential inside the sphere must be constant. Let $C$ be this constant.

It seems that

$$ C = \lim_{r \to R^+} V(r) = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R} $$

But why is this true? My textbook says: because the electric potential must be a continuous function. But why? I am hoping for a non-experimental reason.

Please be precise when mentioning $r<R$ or $r\le R$. Those are different and I get easily confused when people misuse those.

Pedro A
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    Welcome to the site! This is one of the best written "first questions" I have ever seen on this site. Congratulations, and may there be many others. – Floris Apr 12 '15 at 21:50
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    Thank you very much! Hopefully I will also be able to write good answers for other people as well! Thanks! – Pedro A Apr 13 '15 at 02:33
  • Question edited: the equation I first gave for the potential was wrong! @Floris I wonder how you missed it as well. Thankfully this doesn't change the answer for my question. – Pedro A Apr 14 '15 at 11:17

2 Answers2

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Imagine you have a point charge inside the conducting sphere. Obviously, since the electric field inside the sphere is zero (as you state), there is no force on the charge, so no work done. Therefore the potential is constant. So far so good.

Now as we approach the boundary, we can imagine moving an infinitesimal amount to go from $r = R - \delta r$ to $r = R + \delta r$. As long as the electric field is at most some finite amount $E_{shell}$, then the work done moving from just inside to just outside is $E_{shell}*2\delta r$; as $\delta r \rightarrow 0$, the work done will also tend to zero. The only way this would not be true is if the electric field at $r=R$ was infinite - which it is not.

This means that the potential is continuous across the shell, and that in turn means that the potential inside must equal the potential at the surface. Whether we mean by "at the surface" as $R$ or $R + \delta r$ doesn't matter since the difference vanishes as $\delta r$ becomes sufficiently small.

Floris
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  • Thanks! I am getting more and more convinced. But why the electric field is not infinite at r = R? I thought it wasn't defined at all, because the potential isn't differentiable at r = R. – Pedro A Apr 13 '15 at 02:48
  • The finite jump in the field is obtained by Gauss's law - create a "pill box" that crosses the surface of the conductor. On one side the field is zero, on the other it is $\sigma / \epsilon_0$. A finite jump. Infinite gradient but we don't care about that since we need to integrate, not differentiate, to go from $E$ to $V$. – Floris Apr 13 '15 at 10:34
  • Maybe I am getting too philosophical here, but that "pill box" shows that the field right after the surface is finite, and still don't give any information regarding the field on the surface itself, isn't it? By the way I am not just trying to find mistakes in all your explanations, sorry about that. All answers are helping to convince me but I would say I am not 100% there yet... Thanks again. – Pedro A Apr 13 '15 at 16:17
  • I think you are overthinking this. If you make the shell of finite thickness, you can see that the field decreases continuously. As you make the shell of charge thinner, the slope becomes steeper. But at no point does anything allow the electric field to become infinite. – Floris Apr 13 '15 at 16:24
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    Indeed. I calculated the electric field if the shell has a finite thickness, and found out that inside the shell the field increases linearly (approx.), from 0 inside to exactly $\frac{Q}{4\pi\epsilon_0 b^2}$ where $b$ is the outer radius. Perfect - there is no way it is infinite. Thank you very much! – Pedro A Apr 13 '15 at 19:01
  • Glad you got there... it's more satisfying if you can take that last step yourself. – Floris Apr 13 '15 at 19:03
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My textbook says: because the electric potential must be a continuous function. But why?

Since, in the electrostatic case,

$$\vec E(\vec r) = -\nabla V(\vec r)$$

then if the electric field is to be finite everywhere, $V(\vec r)$ must be continuous.

Put less rigorously, the electric field would be 'infinite' wherever $V(\vec r)$ is discontinuous. Since the electric field is observable, we simply can't have that.

Now, the electric field itself can be discontinuous across a boundary.

Community
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Alfred Centauri
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  • If I'm not mistaken, for the gradient to be defined, all partial derivatives must be defined, which is not the case at $r = R$. Therefore, based on the equation you mentioned, the electric field is not defined at $r = R$ (the derivative does not exist), which still leads to my question. – Pedro A Apr 13 '15 at 16:25