Question

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

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

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.

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: \(\Delta V = \frac{Q}{4\pi\epsilon_0\epsilon_r} \left[\frac{1}{a} - \frac{1}{b}\right]\)

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 \(\pi / 2<\phi<2 \pi\).#

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: \(\Delta V = \Delta V_1 + \Delta V_2\)

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: \(\Delta V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{\epsilon_{r1}} + \frac{1}{\epsilon_{r2}}\right) \left[\frac{1}{a} - \frac{1}{b}\right]\)

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 \(C = 8.87 * 10^{-9} \mathrm{F}\).