# Capacitance of Concentric Cylinders

In this topic we will calculate the capacitance of a system of concentric cylindrical shells.

The next figure presents the geometry of this topic. Two conducting and concentric cylindrical shells, infinitely long, constitute a system that some capacitance, C.

Begin by noting that we can only determine the capacitance per unit length, C/L, because the total capacitance is an unreal value. The definition of capacitance is, $C = \frac{Q}{V}$

where Q is the charge of the system and V is its potential. From this expression it is seen that capacitance is the amount of charge that can be stored in a system by holding it at a potential, V. The charge is the amount that can be held separate, not just the total. For example, in a parallel plate capacitor the charge used in $C = Q/V$ is equal to the absolute value of the amount on just one of the plates (if we used the total charge then we would have zero).

To determine the capacitance of this system we will place some charge of the cylinders. Put +Q on the center cylinder and -Q on the outer. The net charge of the system remains zero. This charge will distribute itself over the surface of the cylinders. Our expression for the capacitance per unit length of this system is, $\frac{C}{L} = \frac{Q}{VL}$

where Q = Q is known (we put it there ourselves). It remains to determine the potential, V, that is maintained between the cylinders by the separation of this charge.

Since we know where all the charge is in this system it is possible to determine the electric field everywhere. Knowing the electric field, E, between the cylinders allows for the calculation of the potential through the relation, $V = -\int \vec{E}\cdot d\vec{l}$

where this represents the potential between the end points of the line l. The line taken here will be along the radial coordinate connecting the cylinders. The system is symmetric and this connecting line between the cylinders represents the potential between them at all points.

The symmetry of the system is further exemplified by the electric field pattern of the inner cylinder shown in the figure below. The infinitely long cylinder produces a uniform electric field along the r vector in the cylindrical coordinate system.

Returning to the problem of calculating the electric field, recall Gauss’ law, $\frac{Q_\text{enc}}{\epsilon_0} = \oint \vec{E}\cdot d\vec{A}$

where Qenc is the total charge enclosed by an area A.

Since we want to determine the electric field between the cylinders it is necessary to find a surface that is everywhere perpendicular to it (i.e. a surface with a normal that is parallel to the electric field). The dot product in the above expression is non-zero in such a case. Figure 3 illustrates the surface that satisfies this requirement. The normal vector of the Gaussian surface, A, is everywhere parallel to the electric field vector.

In the above figure, only one example electric field vector has been drawn. This field is everywhere parallel to the radial coordinate vector. The charge of the outer cylinder does not contribute to the total charge enclosed by the surface. The enclosed charge is determined entirely from that of the inner cylinder. This cylindrical surface is three dimensional and represents another cylindrical shell. The entire charge of the innermost cylinder is enclosed by our surface, Qenc = +Q.

The differential area element, dA, can be rewritten in terms of this geometry. The total area of the Gaussian surface is given by the expression for the surface area of a cylinder of radius r. The surface is defined at a fixed radial position, so only the axial (z) and azimuthal (Φ) coordinates are necessary to compute the total area. This is shown below.

Returning to Gauss’ law, let us solve each side separately, $\frac{Q_\text{enc}}{\epsilon_0} = \frac{Q}{\epsilon_0}$ $\oint \vec{E} \cdot d\vec{A} = \oint \left(E_r \hat{r} \right) \cdot \left(r d\phi dz \hat{r} \right)$ $\oint \vec{E} \cdot d\vec{A} = \int^{2\pi}_0 \int^{+\infty}_{-\infty} rE_r d\phi dz$ $\oint \vec{E} \cdot d\vec{A} = 2\pi LE_r r$

where the last step makes use of the fact that I am saying the length of the infinitely long cylinders can be written as L. $\frac{Q}{\epsilon_0} = 2\pi LE_r r$ $E_r = \frac{Q}{2\pi\epsilon_0 Lr}$

where this is directed along the radial coordinate vector. $V = -\int^a_b E_r dr$ $V = -\int^a_b \frac{Q}{2\pi\epsilon_0 Lr} dr$ $V = -\frac{Q}{2\pi\epsilon_0 L} \int^a_b \frac{dr}{r}$ $V = -\frac{Q}{2\pi\epsilon_0 L} \left.\ln(r)\right|^a_b$ $V = -\frac{Q}{2\pi\epsilon_0 L} \left(\ln(a) - \ln(b)\right)$ $V = -\frac{Q}{2\pi\epsilon_0 L} \ln\left( \frac{a}{b}\right)$

where it is very important to remember the reasoning behind the order of the limits in the integral.

Part of the definition of electric potential is that the potential at infinity is zero. When calculating the potential it is necessary to perform the line integration beginning at infinity (or just as far away as possible) and work your way back in. That is why the integration limits proceed from the outermost point, b, and end at the innermost point, a.

The potential is negative, which presents a problem for us because capacitance is positive definite. The following identity is useful, ln(α/β) = -ln(β/α). The potential between the cylinders, after we place equal and opposite amounts of charge on them, is, $V = \frac{Q}{2\pi\epsilon_0 L} \ln\left(\frac{b}{a}\right)$

and we can now return to the earlier generic expression to calculate the capacitance per unit length of this system. $\frac{C}{L} = \frac{Q}{VL}$ $\frac{C}{L} = \frac{Q}{\frac{Q}{2\pi\epsilon_0L} \ln\left(\frac{b}{a} \right)L}$ $\frac{C}{L} = \frac{2\pi\epsilon_0}{\ln\left( \frac{b}{a}\right)}$

where the capacitance per unit length is not a function of either the charge on the system or L itself. The only parameters that matter are the radii of the cylindrical shells. This should be expected because capacitance is a feature of a system’s geometry and does not depend on applied charges or potentials.

## 2 thoughts on “Capacitance of Concentric Cylinders”

1. Abdullah says:

Hi! Nice article! Helped me in my assignment, Thank you!

• Pace says: