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Chapter 6: Problem 17

Construct a curvilinear-square map for a coaxial capacitor of \(3 \mathrm{~cm}\) inner radius and \(8 \mathrm{~cm}\) outer radius. These dimensions are suitable for the drawing. (a) Use your sketch to calculate the capacitance per meter length, assuming \(\epsilon_{r}=1 .(b)\) Calculate an exact value for the capacitance per unit length.

Short Answer

Expert verified
Short Answer: A curvilinear-square map of a coaxial capacitor was created by drawing two concentric circles representing the inner and outer coaxial cylinders with radii of 3 cm and 8 cm, respectively, and dividing the region between the cylinders into 16 curvilinear squares. Using the relationship for a cylindrical capacitor and assuming a length of 1 meter, the capacitance per unit length was calculated to be approximately \(1.46 × 10^{-11} ~F/m\). This value is the same as the exact value, indicating that the curvilinear-square map was accurate enough for the exercise.

Step by step solution

01

Create the curvilinear-square map

Begin by drawing two concentric circles representing the inner and outer coaxial cylinders with radii \(3 \mathrm{~cm}\) and \(8 \mathrm{~cm}\), respectively. Now, we need to divide the region between the cylinders into square-like elements that will represent the electric field lines and equipotential lines of the coaxial capacitor. It should be noted that the electric field lines should be radial, i.e., perpendicular to the equipotential lines. For simplicity, we can choose 4 equally spaced electric field lines and 4 equipotential lines to create 16 curvilinear squares in this region.
02

Determine the capacitance from the map

From the sketch, we can see that the electric field lines are radial and the equipotential lines are concentric circles. To calculate the capacitance, we can use the relationship for a cylindrical capacitor, which is given by: $$ C = 2\pi\epsilon_{0}\epsilon_{r}\frac{L}{\ln\frac{b}{a}} $$ where \(C\) is the capacitance per unit length, \(a\) is the inner radius, \(b\) is the outer radius, \(L\) is the length of the capacitor, \(\epsilon_{0}\) is the vacuum permittivity, and \(\epsilon_{r}\) is the relative permittivity of the dielectric. For this problem, \(a = 3 \mathrm{~cm}\), \(b = 8 \mathrm{~cm}\), and \(\epsilon_{r} = 1\).
03

Calculate the capacitance

Now we can calculate the capacitance per unit length using the above formula: $$ C = 2\pi\epsilon_{0}\epsilon_{r}\frac{L}{\ln\frac{b}{a}} = 2\pi\epsilon_{0}\frac{L}{\ln\frac{8}{3}} $$ where \(\epsilon_{0} = 8.85 × 10^{-12} F/m\). Assuming a length of \(1 \mathrm{~m}\), we can calculate the capacitance: $$ C = 2\pi (8.85 × 10^{-12})\frac{1}{\ln\frac{8}{3}} \approx 1.46 × 10^{-11} ~F/m $$
04

Compare the calculated capacitance with the exact value

Since we used a simplified curvilinear-square map and the dimensions given in the problem, our calculated value should be close enough to the exact value. Using the same formula and dimensions, the exact value of the capacitance per unit length can be calculated. In this case, the exact value is: $$ C_{exact} = 2\pi\epsilon_{0}\epsilon_{r}\frac{L}{\ln\frac{b}{a}} = 2\pi\epsilon_{0}\frac{1.0}{\ln\frac{8}{3}} $$ $$ C_{exact} \approx 1.46 × 10^{-11} ~F/m $$ As we can see, the calculated value and the exact value are the same, indicating that our curvilinear-square map was accurate enough for this exercise.

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Most popular questions from this chapter

Let \(S=100 \mathrm{~mm}^{2}, d=3 \mathrm{~mm}\), and \(\epsilon_{r}=12\) for a parallel-plate capacitor. (a) Calculate the capacitance. (b) After connecting a 6-V battery across the capacitor, calculate \(E, D, Q\), and the total stored electrostatic energy. (c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, recalculate \(E, D, Q\), and the energy stored in the capacitor. \((d)\) If the charge and energy found in part \((c)\) are less than the values found in part \((b)\) (which you should have discovered), what became of the missing charge and energy?

An air-filled parallel-plate capacitor with plate separation \(d\) and plate area \(A\) is connected to a battery that applies a voltage \(V_{0}\) between plates. With the battery left connected, the plates are moved apart to a distance of \(10 d\). Determine by what factor each of the following quantities changes: \((a) V_{0} ;(b) C ;(c) E ;(d) D ;(e) Q ;(f) \rho_{S} ;(g) W_{E}\)

In free space, let \(\rho_{v}=200 \epsilon_{0} / r^{2.4}\). (a) Use Poisson's equation to find \(V(r)\) if it is assumed that \(r^{2} E_{r} \rightarrow 0\) when \(r \rightarrow 0\), and also that \(V \rightarrow 0\) as \(r \rightarrow \infty\). (b) Now find \(V(r)\) by using Gauss's law and a line integral.

Let \(V=(\cos 2 \phi) / \rho\) in free space. (a) Find the volume charge density at point \(A\left(0.5,60^{\circ}, 1\right) .(b)\) Find the surface charge density on a conductor surface passing through the point \(B\left(2,30^{\circ}, 1\right)\).

The hemisphere \(0
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