In deriving the capacitance for a cylindrical and spherical capacitors, I keep obtaining the incorrect sign on V. I completely understand the problem besides one step.

First we start with Gauss' law:

$ε_0φ = q_{enc}$

$ε_0\int{E • dA} = q_{enc}$

if we were to draw a cylindrical gaussian surface concentric with the two cylinders, the angle between the vector of the area of the surface and the net electric field would be 0. Therefore, the dot product would be 1. Also, E would be constant on the surface due to symmetry, so E can be taken out of the integral leaving only dA. Integrating dA yields A:

$ε_0EA = q_{enc}$

$ε_0E(2πrL) = q_{enc}$ where r is the radius of the cylindrical surface and L is the length of the surface.

$E = \frac{q_{enc}}{2πrLε_0}$

Now we substitute this in for E into:

$V = -\int_R^r{E • dS}$

where R is the radius of the larger Cylinder and r is the radius of the smaller cylinder. We integrate by moving a positive test charge from the negative to positive cylinder (assuming the inner cylinder is positive); therefore, V should be positive.

$V = -\int_R^r{\frac{q_{enc}}{2πrLε_0} • dS}$

the dot product between the electric field and dS would be -1 if we consider the angle between the vectors to be 180.

$V = \int_R^r{\frac{q_{enc}}{2πrLε_0} dS}$

substitute in dr=dS

$V = \int_R^r{\frac{q_{enc}}{2πrLε_0} dr}$

$V = \frac{q_{enc}}{2πLε_0} [ln|r| + C]_R^r$

$V = \frac{q_{enc}}{2πLε_0}ln(r/R)$

more in comments

  • $\begingroup$ however, since r<R, ln(r/R) would yield a negative value and therefore a negative V. Where is my error? The textbook says that the substitution for dS is dS = -dR, but that's the step I don't understand. Is dS automatically assumed to be going in the direction of the electric field, so to integrate radially inward we would need to delegate a sign change? Or could it be that the domain of the dot product is from 0 to 90, so I factored in an extra negative by considering an angle of 180 degrees instead of 0. Any help would be greatly appreciated! $\endgroup$
    – Lightbulb
    Feb 20 '20 at 14:49
  • $\begingroup$ The voltage drops going in the direction of the field. Your answer will depend on your choice of limits. $\endgroup$
    – R.W. Bird
    Feb 20 '20 at 20:32

If you integrate from the negative plate to the positive plate the potential difference is positive (+V). You have V is the negative integral of E, which is true if you integrate in the direction of E (E dot dS)


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