I'm trying to solve a doubt that has taken me away way too much time and so I'm asking here. In our lectures we defined the electric potential difference between a point A and B as the line integral over some path connecting the two points, of the electric field:

$$\Delta V = V_B - V_A = \int\limits_B^A\vec{E}\cdot d\vec{s} = -\int\limits_A^B\vec{E}\cdot d\vec{s}$$

Now I wanna find the potential difference between the plates of a parallel plates capacitor and therefore I take a point on the lower plate (I like thinking of them as vertical rather than horizontal) and a point on the upper plate, to make things easier I take them vertically aligned and I define a vertical y axis pointing upwards and with its origin in corrispondence of the lower plate, so that the y coordinate of the upper plate is y = d. According to that definition if I decide to start from the point A on the lower one and B on the upper one I find that (if +Q is the charge I put on the lower plate)

$$\Delta V = V_d - V_0 = -\int_\Gamma \vec{E}\cdot\vec{ds} = -\int_\Gamma \frac{Q}{\epsilon_0 S}{\widehat{u}_y}\cdot{{\widehat{u}_y}} = -\frac{Q}{\epsilon_0 S}\int\limits_0^d dy = \frac{Q}{\epsilon_0 S}\int\limits_d^0 dy = -\frac{Qd}{\epsilon_0 S}$$

And therefore the capacitance would be negative. This also keeps being true if I repeat the same process with a cylindrical capacitor: if I call R₁ the radius of the external cylinder and R₂ the radius of the internal one, I put some +Q charge on the inner one and then want to calculate the electric potential difference between a point A on the surface of the internal one and a point B on the surfare of the outer one I get that

$$\Delta V = V_{R_2} - V_{R_1} = -\int_\Gamma \vec{E}\cdot\vec{ds} = -\int\limits_{R_1}^{R_2} \frac{Q}{2\pi Rh\epsilon_0}{\widehat{u}_R}\cdot dR{{\widehat{u}_R}} = \frac{Q}{2\pi h\epsilon_0} \int\limits_{R_2}^{R_1} \frac{1}{R}dR = \frac{Q}{2\pi h\epsilon_0}\ln(\frac{R_1}{R_2})$$

And again it comes out to be negative since R₁ < R₂. My question is: do I need to take the electric potential difference positive independently of the sign of the result I got? Or maybe should I always start from the negatively charged plate of my capacitor to get a positive result? (I'm pretty sure I can't get negative capacitance).


1 Answer 1


In the definition of capacitance $C=Q/V$, the voltage $V$ is just the magnitude of the potential difference. You don't worry about the sign. This is because $Q$ also is just a magnitude (you have positive charge on one plate and negative charge on another plate). So there is no issue here.

Certainly you are correct in your work. In moving from the positive to the negative plate the potential should decrease.


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