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Trivial question about electric potential

  1. Aug 20, 2014 #1
    Hi! I'm having an hard time with a trivial question that suddenly I can't figure out: Computing the potential between the plates of a spherical capacitor.

    The problem is taken from this url (first page and a half): http://www2.warwick.ac.uk/fac/sci/p...e_home/px263/assignments/assignment2_soln.pdf

    Imagine a spherical capacitor in which the inner surface has radius [itex]a[/itex] and the outer surface radius [itex]b[/itex], so [itex]a<b[/itex]. The inner surface has a total charge of [itex]Q[/itex] and he outer surface has a total charge of [itex]-Q[/itex]. The electric field can be easily computed by Gauss's law and points radially outwards; the potential between the plates is computed from it using

    [itex] V(\vec{r})=-\intop_{O}^{\vec{r}}\overrightarrow{E}\cdot d\overrightarrow{l} [/itex]


    If you check the document you can see they do the calculation in the path from [itex]r=b[/itex] to [itex]r=a[/itex], obtaining the potential difference between plates

    [itex] V(b)-V(a)=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{a}-\frac{1}{b})>0\Leftrightarrow V(b)>V(a) [/itex]

    Is this correct? We know the eletric field points in the direction of decreasing potential but, from this example it is pointing in the direction of increasing potential. Why is this?

    I was doing the integration going from [itex]r=a[/itex] to [itex]r=b[/itex] and I correctly (in my mind) get [itex]V(b)<V(a)[/itex]. Since we're moving in the positive direction of [itex]\hat{r}[/itex] the dot product with [itex]\vec{E}[/itex] will be positive at all times, because [itex]\vec{E}\cdot d\overrightarrow{l}=E\hat{r}\cdot dr\hat{r}=Edr[/itex]

    [itex] V(b)-V(a)=-\intop_{a}^{b}\overrightarrow{E}\cdot d\overrightarrow{l}=-\intop_{a}^{b}\frac{Q}{4\pi\epsilon_{0}r^{2}}dr=\frac{Q}{4\pi\epsilon_{0}}(\frac{1}{b}-\frac{1}{a})<0\Leftrightarrow V(b)<V(a) [/itex]

    But the thing is, if I try to do it from [itex]r=b[/itex] to [itex]r=a[/itex] as in that solution sheet, I get the opposite, just like them.

    [itex] V(a)-V(b)=-\intop_{b}^{a}\overrightarrow{E}\cdot d\overrightarrow{l}=-\intop_{b}^{a}-\frac{Q}{4\pi\epsilon_{0}r^{2}}dr=\frac{Q}{4\pi\epsilon_{0}}(\frac{1}{b}-\frac{1}{a})<0\Leftrightarrow V(b)>V(a) [/itex]

    So, where is the confusion? Which way is right and why?
     
  2. jcsd
  3. Aug 20, 2014 #2

    TSny

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    Hello, ppedro. The mistake is in writing ##d\vec{l} = -dr\;\hat{r}## when integrating from b to a. The negative sign should not be there.

    You should always write ##d\vec{l} = dr\;\hat{r}## whether you integrate outward along a radius from a to b or inward along a radius from b to a.

    When integrating outward, the infinitesimal change in r, dr, is positive which makes ##d\vec{l} = dr\;\hat{r}## point outward, as it should.

    When integrating inward from b to a, dr is negative. Then ##d\vec{l} = dr\;\hat{r}## points inward, as it should.
     
  4. Aug 20, 2014 #3
    The most common confusion students have when they first study electric potential .Your analysis is spot on :smile:
     
  5. Aug 21, 2014 #4
    Thanks TSny! It makes sense. But that would mean the answer in the solution sheet I shared is wrong, right?
     
  6. Aug 21, 2014 #5

    TSny

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    Yes, the solution sheet is wrong.
     
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