Question

Two conducting spherical shells have radii $a=3\mathrm{cm}$ and $b=6\mathrm{cm}.$ The interior is a perfect dielectric for which ${ϵ}_r=8$. (a) Find $C.(b)$ A portion of the dielectric is now removed so that ${ϵ}_r=1.0,0<\varphi <\pi /2$, and ${ϵ}_r=8$, $\pi /2<\varphi <2\pi$. 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 ${ϵ}_r=8$, the capacitance is $C=8.91\ast {10}^{-9}\mathrm{F}$. In the case when the dielectric constant varies depending on the angle, with ${ϵ}_r=1.0$ for $0<\varphi <\pi /2$ and ${ϵ}_r=8$ for $\pi /2<\varphi <2\pi$, the capacitance is $C=8.87\ast {10}^{-9}\mathrm{F}$.

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: $E=\frac{Q}{4\pi {ϵ}_0{ϵ}_r{r}^2}$ where $Q$ is the charge on the spheres, ${ϵ}_0$ is vacuum permittivity, ${ϵ}_r$ 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 $b$: $\mathrm{\Delta }V=-{\int }_a^{b}Edr=\frac{Q}{4\pi {ϵ}_0{ϵ}_r}{\int }_a^b\frac{1}{r^2}dr$ Evaluate the integral and we get: $\mathrm{\Delta }V=\frac{Q}{4\pi {ϵ}_0{ϵ}_r}[\frac{1}{a}-\frac{1}{b}]$

Calculate the capacitance.

Now, we can find the capacitance using the formula $C=\frac{Q}{\mathrm{\Delta }V}$: $C=\frac{Q\ast (4\pi {ϵ}_0{ϵ}_r)}{Q\ast (\frac{1}{a}-\frac{1}{b})}=4\pi {ϵ}_0{ϵ}_r\frac{ab}{b-a}$ Plug the values of $a$, $b$, ${ϵ}_0$, and ${ϵ}_r$: $C=4\pi \ast 8.85\ast {10}^{-12}\ast 8\ast \frac{3\ast {10}^{-2}\ast 6\ast {10}^{-2}}{3\ast {10}^{-2}}=4\pi \ast 8.85\ast {10}^{-10}=8.91\ast {10}^{-9}\mathrm{F}$ So, the capacitance in the case when ${ϵ}_r=8$ is $C=8.91\ast {10}^{-9}\mathrm{F}$. #b. Find $C$ when ${ϵ}_r=1.0$ for $0<\varphi <\pi /2$ and ${ϵ}_r=8$ for $\pi /2<\varphi <2\pi$.#

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 $\varphi$. We need to divide the space in two regions with ${ϵ}_{r1}=1$ ($0<\varphi <\pi /2$) and ${ϵ}_{r2}=8$ ($\pi /2<\varphi <2\pi$). To find the total potential difference, we have to sum the potential differences for both regions: $\mathrm{\Delta }V=\mathrm{\Delta }{V}_1+\mathrm{\Delta }{V}_2$

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: $\mathrm{\Delta }{V}_1=\frac{Q}{4\pi {ϵ}_0{ϵ}_{r1}}[\frac{1}{a}-\frac{1}{b}]$ $\mathrm{\Delta }{V}_2=\frac{Q}{4\pi {ϵ}_0{ϵ}_{r2}}[\frac{1}{a}-\frac{1}{b}]$ Sum these two to get the total potential difference: $\mathrm{\Delta }V=\frac{Q}{4\pi {ϵ}_0}(\frac{1}{{ϵ}_{r1}}+\frac{1}{{ϵ}_{r2}})[\frac{1}{a}-\frac{1}{b}]$

Calculate the capacitance.

Use the formula $C=\frac{Q}{\mathrm{\Delta }V}$ again: $C=\frac{Q}{\frac{Q}{4\pi {ϵ}_0}(\frac{1}{{ϵ}_{r1}}+\frac{1}{{ϵ}_{r2}})(\frac{1}{a}-\frac{1}{b})}=4\pi {ϵ}_0\frac{{ϵ}_{r1}{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}\frac{ab}{b-a}$ Plug the values of $a$, $b$, ${ϵ}_0$, ${ϵ}_{r1}$, and ${ϵ}_{r2}$: $C=4\pi \ast 8.85\ast {10}^{-12}\ast \frac{8}{1+8}\ast \frac{3\ast {10}^{-2}\ast 6\ast {10}^{-2}}{3\ast {10}^{-2}}=4\pi \ast 7.075\ast {10}^{-10}=8.87\ast {10}^{-9}\mathrm{F}$ So, the capacitance in the case when ${ϵ}_r=1.0$ for $0<\varphi <\pi /2$ and ${ϵ}_r=8$ for $\pi /2<\varphi <2\pi$ is $C=8.87\ast {10}^{-9}\mathrm{F}$.

Key concepts

Dielectric Material

Dielectric materials are insulating substances that can store electrical energy by polarizing when exposed to an electric field. They don't conduct electricity; instead, they increase a capacitor's ability to store charge. This is quantified through the relative permittivity, known as the dielectric constant (${ϵ}_r$).

• The dielectric constant represents how much electric field strength is reduced within the material compared to a vacuum.
• High dielectric constants help in storing more charge, hence increasing the capacitance.
• In our example, the region between the spherical shells initially has a uniform dielectric with ${ϵ}_r=8$, boosting the capacitance significantly.

Understanding and choosing the right dielectric material is crucial in designing efficient capacitors in electronic circuits.

Electric Field

An electric field is a region around a charged object where the object's electric force is felt by other charges. It describes the force per unit charge exerted on a positively charged test particle. For spherical shells, the electric field $E$ can be given by:$$E=\frac{Q}{4\pi {ϵ}_0{ϵ}_r{r}^2}$$

• Here, $Q$ represents charge, and $r$ is the distance from the center of the spheres.
• The field strength decreases with the square of the distance ($1/r^2$), indicating it's stronger near the charges and weaker at points farther away.
• In the presence of a dielectric, the field is reduced by a factor of ${ϵ}_r$, diminishing the force felt by any test charges.

This reduction allows more charge to be stored for the same potential, contributing to increased capacitance.

Potential Difference

The potential difference between two points is the work needed to move a unit charge from one point to another. In a capacitor, it represents the energy per charge stored and can be calculated by integrating the electric field over the distance. For spherical shells, the potential difference $\mathrm{\Delta }V$ is given by:$$\mathrm{\Delta }V=\frac{Q}{4\pi {ϵ}_0{ϵ}_r}[\frac{1}{a}-\frac{1}{b}]$$

• This shows how potential difference depends on the dielectric material (${ϵ}_r$) and the geometry ($a,b$) of the shells.
• A higher potential difference indicates more energy stored per charge.
• When dielstric properties vary, as in part (b) of the exercise, calculating potential differences for different regions is crucial.

By understanding this concept, you grasp the basis for energy storage in capacitors and the effect of dielectrics on them.

Conducting Spherical Shells

Conducting spherical shells are often used in physics as a model for capacitors. They consist of two spherical conductors: one inside the other, separated by an insulating material. Important features include:

• The conducting properties ensure a uniform charge distribution on each shell.
• They allow for the analysis of electric fields and potential differences in a simplified, symmetrical setup.
• The capacitance of such systems depends heavily on the radii of the spheres and the dielectric material between them.

In our exercise, finding the capacitance involves understanding how these shells and the dielectric in between affect electrical properties. Knowing how to calculate these factors in spherical geometry is essential for various applications in electronics and physics.