Trivial sign problem Please set me straight

I'm calculating the capacitance of a set of spherical shells of radii b > a. To do this, I place a charge +Q on the inner shell and -Q on the outer shell so that I get the electric field vector pointing outward
[tex]
\vec E = \frac{Q}{4\pi\varepsilon_0} \frac{\hat r}{r^2}.
[/tex]

Finding the potential should be trivial... I do the integral
[tex]
V = -\int_a^b \vec E \cdot d\vec r = -\frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = -\frac{Q}{4\pi\varepsilon_0}\left( \frac{1}{a} - \frac{1}{b} \right).
[/tex]

This is off by a minus sign and I cannot understand why... reversing the limits of integrations, of course, kills the sign but introduces a sign flip in the dot product as E then opposes dr. This should be trivial... help? :)
 

berkeman

Mentor
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I'm calculating the capacitance of a set of spherical shells of radii b > a. To do this, I place a charge +Q on the inner shell and -Q on the outer shell so that I get the electric field vector pointing outward
[tex]
\vec E = \frac{Q}{4\pi\varepsilon_0} \frac{\hat r}{r^2}.
[/tex]

Finding the potential should be trivial... I do the integral
[tex]
V = -\int_a^b \vec E \cdot d\vec r = -\frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = -\frac{Q}{4\pi\varepsilon_0}\left( \frac{1}{a} - \frac{1}{b} \right).
[/tex]

This is off by a minus sign and I cannot understand why... reversing the limits of integrations, of course, kills the sign but introduces a sign flip in the dot product as E then opposes dr. This should be trivial... help? :)
Are you sure there is a sign error? If you have +Q at a and -Q at b, which way does the Electric field point?


EDIT -- If you are moving in the direction of the E-field, your potential is decreasing. Still seems like the signs are correct to me.
 
It would have to point from =Q to -Q (outward radial), no? That's why I'm confused... I could take the absolute value to get the correct value for the capacitance but still... maybe it's confusion in calculating capacitance? As best practice should one always do the integration against the field (which would ensure an increase in V) ?

Thanks!

EDIT: In studying for finals I am comparing to Jackson solutions... most people glibly neglect the minus sign out front to get the right answer, whilst some other souls take the absolute value. It seems better to me to simply integrate against the field?
 

berkeman

Mentor
54,317
4,722
It would have to point from =Q to -Q (outward radial), no? That's why I'm confused... I could take the absolute value to get the correct value for the capacitance but still... maybe it's confusion in calculating capacitance? As best practice should one always do the integration against the field (which would ensure an increase in V) ?

Thanks!

EDIT: In studying for finals I am comparing to Jackson solutions... most people glibly neglect the minus sign out front to get the right answer, whilst some other souls take the absolute value. It seems better to me to simply integrate against the field?
Yes, I'm inclined to integrate against the field as well. After all, C = Q/V, and to get a +Q, you need to integrate from the -Q to the +Q location. Going that direction also gives you a +V, hence a +C. I suppose you can to the other direction by symmetry, and get a -Q/-V to get the same +C.
 
Great, thanks! It's things like this that I really should have nailed down by now... but better to get it now than right before the written exam, I suppose.
 

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