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Chapter 6: Problem 11

Two conducting spherical shells have radii \(a=3 \mathrm{~cm}\) and \(b=6 \mathrm{~cm} .\) The interior is a perfect dielectric for which \(\epsilon_{r}=8\). (a) Find \(C .(b)\) A portion of the dielectric is now removed so that \(\epsilon_{r}=1.0,0<\phi<\pi / 2\), and \(\epsilon_{r}=8\), \(\pi / 2<\phi<2 \pi\). Again find \(C\).

Short Answer

Expert verified
In summary, for the case when the region between the spherical shells is uniformly filled with a dielectric constant of \(\epsilon_{r}=8\), the capacitance is \(C = 8.91 * 10^{-9} \mathrm{F}\). In the case when the dielectric constant varies depending on the angle, with \(\epsilon_{r}=1.0\) for \(0<\phi<\pi/2\) and \(\epsilon_{r}=8\) for \(\pi/2<\phi<2\pi\), the capacitance is \(C = 8.87 * 10^{-9} \mathrm{F}\).

Step by step solution

01

Calculate the electric field between the spheres.

Since the dielectric fills the region between the spheres uniformly, the electric field between the spheres is given by: \(E = \frac{Q}{4\pi\epsilon_0\epsilon_r r^2}\) where \(Q\) is the charge on the spheres, \(\epsilon_0\) is vacuum permittivity, \(\epsilon_r\) is relative permittivity, and \(r\) is the distance from the center of the spheres.
02

Get the potential difference between the spheres.

The potential difference between the spheres is equal to the work done in moving a positive test charge from the outer sphere to the inner sphere against the electric field. We can find the potential difference by integrating the electric field over the distance from \(a\) to \(b\): \(\Delta V = -\int_a^b E dr = \frac{Q}{4\pi\epsilon_0\epsilon_r} \int_a^b \frac{1}{r^2} dr\) Evaluate the integral and we get: \(\Delta V = \frac{Q}{4\pi\epsilon_0\epsilon_r} \left[\frac{1}{a} - \frac{1}{b}\right]\)
03

Calculate the capacitance.

Now, we can find the capacitance using the formula \(C=\frac{Q}{\Delta V}\): \(C = \frac{Q*(4\pi\epsilon_0\epsilon_r)}{Q*(\frac{1}{a} - \frac{1}{b})} = 4\pi\epsilon_0\epsilon_r \frac{ab}{b - a}\) Plug the values of \(a\), \(b\), \(\epsilon_0\), and \(\epsilon_r\): \(C = 4\pi * 8.85 * 10^{-12} * 8 * \frac{3 * 10^{-2} * 6 * 10^{-2}}{3 * 10^{-2}} = 4\pi * 8.85 * 10^{-10} = 8.91 * 10^{-9} \mathrm{F}\) So, the capacitance in the case when \(\epsilon_{r}=8\) is \(C = 8.91 * 10^{-9}\mathrm{F}\). #b. Find \(C\) when \(\epsilon_{r}=1.0\) for \(0<\phi<\pi / 2\) and \(\epsilon_{r}=8\) for \(\pi / 2<\phi<2 \pi\).#
04

Divide the space in two regions and find the total potential difference between the spheres.

In this case, the dielectric constant varies depending on the angle \(\phi\). We need to divide the space in two regions with \(\epsilon_{r1} = 1\) (\(0<\phi<\pi / 2\)) and \(\epsilon_{r2} = 8\) (\(\pi / 2<\phi<2 \pi\)). To find the total potential difference, we have to sum the potential differences for both regions: \(\Delta V = \Delta V_1 + \Delta V_2\)
05

Calculate the potential differences for each region and sum them.

Using the integral and the electric field formula for each region, we find the potential differences: \(\Delta V_1 = \frac{Q}{4\pi\epsilon_0\epsilon_{r1}} \left[\frac{1}{a} - \frac{1}{b}\right]\) \(\Delta V_2 = \frac{Q}{4\pi\epsilon_0\epsilon_{r2}} \left[\frac{1}{a} - \frac{1}{b}\right]\) Sum these two to get the total potential difference: \(\Delta V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{\epsilon_{r1}} + \frac{1}{\epsilon_{r2}}\right) \left[\frac{1}{a} - \frac{1}{b}\right]\)
06

Calculate the capacitance.

Use the formula \(C=\frac{Q}{\Delta V}\) again: \(C = \frac{Q}{\frac{Q}{4\pi\epsilon_0} (\frac{1}{\epsilon_{r1}} + \frac{1}{\epsilon_{r2}}) (\frac{1}{a} - \frac{1}{b})} = 4\pi\epsilon_0\frac{\epsilon_{r1}\epsilon_{r2}}{\epsilon_{r1}+\epsilon_{r2}} \frac{ab}{b - a}\) Plug the values of \(a\), \(b\), \(\epsilon_0\), \(\epsilon_{r1}\), and \(\epsilon_{r2}\): \(C = 4\pi * 8.85 * 10^{-12} * \frac{8}{1+8} * \frac{3 * 10^{-2} * 6 * 10^{-2}}{3 * 10^{-2}} = 4\pi * 7.075 * 10^{-10} = 8.87 * 10^{-9} \mathrm{F}\) So, the capacitance in the case when \(\epsilon_{r}=1.0\) for \(0<\phi<\pi/2\) and \(\epsilon_{r}=8\) for \(\pi/2<\phi<2\pi\) is \(C = 8.87 * 10^{-9} \mathrm{F}\).

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Most popular questions from this chapter

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 \(B\left(2,30^{\circ}, 1\right)\).

A parallel-plate capacitor has plates located at \(z=0\) and \(z=d\). The region between plates is filled with a material that contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\) and has permittivity \(\epsilon\). Both plates are held at ground potential. ( \(a\) ) Determine the potential field between plates. (b) Determine the electric field intensity \(\mathbf{E}\) between plates. \((c)\) Repeat parts \((a)\) and \((b)\) for the case of the plate at \(z=d\) raised to potential \(V_{0}\), with the \(z=0\) plate grounded.

A parallel-plate capacitor is filled with a nonuniform dielectric characterized by \(\epsilon_{r}=2+2 \times 10^{6} x^{2}\), where \(x\) is the distance from one plate in meters. If \(S=0.02 \mathrm{~m}^{2}\) and \(d=1 \mathrm{~mm}\), find \(C\).

Capacitors tend to be more expensive as their capacitance and maximum voltage \(V_{\max }\) increase. The voltage \(V_{\max }\) is limited by the field strength at which the dielectric breaks down, \(E_{B D}\). Which of these dielectrics will give the largest \(C V_{\max }\) product for equal plate areas? \((a)\) Air: \(\epsilon_{r}=1\), \(E_{B D}=3 \mathrm{MV} / \mathrm{m} .(b)\) Barium titanate: \(\epsilon_{r}=1200, E_{B D}=3 \mathrm{MV} / \mathrm{m} .(c)\) Silicon dioxide: \(\epsilon_{r}=3.78, E_{B D}=16 \mathrm{MV} / \mathrm{m} .(d)\) Polyethylene: \(\epsilon_{r}=2.26, E_{B D}=\) \(4.7 \mathrm{MV} / \mathrm{m} .\)

A coaxial cable has conductor dimensions of \(a=1.0 \mathrm{~mm}\) and \(b=2.7 \mathrm{~mm}\). The inner conductor is supported by dielectric spacers \(\left(\epsilon_{r}=5\right)\) in the form of washers with a hole radius of \(1 \mathrm{~mm}\) and an outer radius of \(2.7 \mathrm{~mm}\), and with a thickness of \(3.0 \mathrm{~mm}\). The spacers are located every \(2 \mathrm{~cm}\) down the cable. ( \(a\) ) By what factor do the spacers increase the capacitance per unit length? \((b)\) If \(100 \mathrm{~V}\) is maintained across the cable, find \(\mathbf{E}\) at all points.
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