Question 24.4: A CYLINDRICAL CAPACITOR Two long, coaxial cylindrical conduc...

A CYLINDRICAL CAPACITOR

Two long, coaxial cylindrical conductors are separated by vacuum (Fig. 24.6). The inner cylinder has radius r_a and linear charge density +λ. The outer cylinder has inner radius r_B and linear charge density -λ. Find the capacitance per unit length for this capacitor.

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IDENTIFY and SET UP:

As in Example 24.3, we use the definition of capacitance, C = Q/V_{ab}. We use the result of Example 23.10 (Section 23.3) to find the potential difference V_{ab} between the cylinders, and find the charge Q on a length L of the cylinders from the linear charge density. We then find the corresponding capacitance C from Eq. (24.1). Our target variable is this capacitance divided by L.

C=\frac{Q}{V_{a b}}                            (24.1)

EXECUTE:

As in Example 24.3, the potential V between the cylinders is not affected by the presence of the charged outer cylinder.
Hence our result in Example 23.10 for the potential outside a charged conducting cylinder also holds in this example for potential in the space between the cylinders:

V=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{r_{0}}{r}

Here r_0 is the arbitrary, finite radius at which V = 0. We take r_0 = r_b, the radius of the inner surface of the outer cylinder. Then the potential at the outer surface of the inner cylinder (at which r = r_a) is just the potential V_{ab} of the inner (positive) cylinder a with respect to the outer (negative) cylinder b:

V_{a b}=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{r_{b}}{r_{a}}

If λ is positive as in Fig. 24.6, then V_{ab} is positive as well: The inner cylinder is at higher potential than the outer.
The total charge Q in a length L is Q = λL, so from Eq. (24.1) the capacitance C of a length L is

C=\frac{Q}{V_{a b}}=\frac{\lambda L}{\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{r_{b}}{r_{a}}}=\frac{2 \pi \epsilon_{0} L}{\ln \left(r_{b} / r_{a}\right)}

The capacitance per unit length is

\frac{C}{L}=\frac{2 \pi \epsilon_{0}}{\ln \left(r_{b} / r_{a}\right)}

Substituting \epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}=8.85 \mathrm{pF} / \mathrm{m}, we get

\frac{C}{L}=\frac{55.6 \mathrm{pF} / \mathrm{m}}{\ln \left(r_{b} / r_{a}\right)}

 

EVALUATE: The capacitance of coaxial cylinders is determined entirely by their dimensions, just as for parallel-plate and spherical capacitors. Ordinary coaxial cables are made like this but with an insulating material instead of vacuum between the conductors.

A typical cable used for connecting a television to a cable TV feed has a capacitance per unit length of 69 pF/m.

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