Trivial question about electric potential

[ppedro]

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/physics/current/teach/module_home/px263/assignments/assignment2_soln.pdf

Imagine a spherical capacitor in which the inner surface has radius $a$ and the outer surface radius $b$, so $a<b$. The inner surface has a total charge of $Q$ and he outer surface has a total charge of $-Q$. 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

$V(\vec{r})=-\intop_{O}^{\vec{r}}\overrightarrow{E}\cdot d\overrightarrow{l}$If you check the document you can see they do the calculation in the path from $r=b$ to $r=a$, obtaining the potential difference between plates

$V(b)-V(a)=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{a}-\frac{1}{b})>0\Leftrightarrow V(b)>V(a)$

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

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

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

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

So, where is the confusion? Which way is right and why?

 

[TSny]

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.

 

[Tanya Sharma]

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.

The most common confusion students have when they first study electric potential .Your analysis is spot on

 

[ppedro]

Thanks TSny! It makes sense. But that would mean the answer in the solution sheet I shared is wrong, right?

 

[TSny]

Yes, the solution sheet is wrong.