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I've got a simple question regarding electric potentials in electrostatics.

My book defines electric potential V as

$$ V(b) = -\int_a^b \mathbf{E}\cdot d\mathbf{l} $$

where A is defined as where potential is zero, usually some infinitely far distance away.

However, I'm getting a little confused in how to apply this equation. Consider a cylindrical wire of charge of radius A and charge per unit length λ; it's at the center of a thin cylindrical conducting shell of radius B . Here's a picture.

enter image description here

To find the capacitance per unit length, I try

$$ \frac{\lambda}{V} = \frac{\lambda}{-\int_a^b \mathbf{E} dr} = \frac{\lambda}{-\int_a^b \frac{\lambda}{2\pi r\epsilon_0} dr} = -\frac{2\pi\epsilon_0}{\log\left|\frac{b}{a}\right|} $$

Obviously the minus sign is incorrect. My book just states to drop the minus sign, but I don't really understand why. Can anyone help? Thanks!

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  • $\begingroup$ The capacitance is defined as positive. In the defining equation $Q=C\Delta V$, the neither the charge $Q$ nor the potential difference $\Delta V$ should be signed. $\endgroup$ – ZeroTheHero Sep 30 '18 at 20:13
  • $\begingroup$ I understand that. My question is that, if V is defined to contain that negative sign, why is it dropped in this particular instance? It would make sense if the potential is zero at B, you could write a negative integral from b to a, which becomes a positive integral from a to b... but I don't think the potential is zero at b?? $\endgroup$ – Programmer Sep 30 '18 at 20:35
  • $\begingroup$ Perhaps a better definition of capacitance would cure your woes: $C=|Q/V|$. $\endgroup$ – Trevor Kafka Sep 30 '18 at 20:38
  • $\begingroup$ The issue is that the “V” you want to use $C=Q/V$ is not the integral, but it’s absolute value. $\endgroup$ – ZeroTheHero Sep 30 '18 at 20:41
  • $\begingroup$ I understand the physical reasons, but I'm trying to understand mathematically where my sign conventions are wrong. I have a vector pointed the wrong way somewhere. Farcher's answer was what I'm looking for. $\endgroup$ – Programmer Sep 30 '18 at 20:57
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Your error is assigning a magnitude to the vector $\vec E$ rather than the component of the electric field in the $\hat r$ direction which is outwards from the centre of the conducting wire.

If the outer conductor has the positive charge then the electric field is in the $\left (-\hat r\right )$ direction.

$$V_{\text{b relative to a}}= -\int_a^b \vec E\cdot d\vec r = -\int_a^b \frac{\lambda}{2\pi r \epsilon_0}\left(-\hat r \right)\cdot dr \, \hat r = \int_a^b \frac{\lambda}{2\pi r \epsilon_0}\, dr $$

which will be a positive quantity.

Update in response to comments made by @AaronStevens and @ZeroTheHHero

My intention was to get the sign of the potential difference correct as was asked by @Programmer.

If the central wire was positive and one wanted to find the potential of the wire relative to that of the cylinder one would have found the same value.

I did not go on to discuss the equations $C = \frac QV$ and $C = \frac {\Delta Q}{\Delta V}$.

The second equation is hinted at in the Wikipedia article Capacitance with the statement

. . . . self-capacitance, which is the amount of electric charge that must be added to an isolated conductor to raise its electric potential by one unit (i.e. one volt) . . . . . The reference point for this potential is a theoretical hollow conducting sphere, of infinite radius, with the conductor centered inside this sphere.

A source states that the charge $Q$ is a positive quantity and is the amount of positive charge which has been transferred from one conductor (or infinity) to the other conductor and $\Delta V$ is the potential difference created between the conductors and to make absolutely certain about signs has the equation as $Q = C|\Delta V|$.

Another source writes

A key thing to bear in mind when using this formula $Q=CV$ is that $Q$ refers to the charge separation of the system. In other words, when I say that a capacitor is charged up to some level $Q$ I mean that I have put $+ Q$ on one part of the capacitor, $-Q$ on the other.

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  • $\begingroup$ Thank you for the mathematical explanation. Seeing the signs in the R direction helped me develop a better understanding. $\endgroup$ – Programmer Sep 30 '18 at 20:56
  • $\begingroup$ If the positive charge is on the inside then your potential difference is negative though. Technically for capacitance you use the magnitude of the potential difference. The problem doesn't even give the sign of $\lambda$ $\endgroup$ – Aaron Stevens Sep 30 '18 at 22:13
  • $\begingroup$ As a point of clarification: is the sign of the charges on the various surfaces of the cap obvious from the question? If you have $+$ charges on the outside surface, shouldn’t you have $\lambda$ negative? Alternatively, does $\lambda$ refer to the (signed) charge on the inside or outside surface? $\endgroup$ – ZeroTheHero Oct 1 '18 at 0:48

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