# Electric potential difference between capacitor's plates, doubt about the sign?

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).

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.