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Two conducting spherical shells have radii a=3 cm and b=6 cm. The interior is a perfect dielectric for which ϵr=8. (a) Find C.(b) A portion of the dielectric is now removed so that ϵr=1.0,0<ϕ<π/2, and ϵr=8, π/2<ϕ<2π. 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 ϵr=8, the capacitance is C=8.91109F. In the case when the dielectric constant varies depending on the angle, with ϵr=1.0 for 0<ϕ<π/2 and ϵr=8 for π/2<ϕ<2π, the capacitance is C=8.87109F.

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=Q4πϵ0ϵrr2 where Q is the charge on the spheres, ϵ0 is vacuum permittivity, ϵr 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: ΔV=abEdr=Q4πϵ0ϵrab1r2dr Evaluate the integral and we get: ΔV=Q4πϵ0ϵr[1a1b]
03

Calculate the capacitance.

Now, we can find the capacitance using the formula C=QΔV: C=Q(4πϵ0ϵr)Q(1a1b)=4πϵ0ϵrabba Plug the values of a, b, ϵ0, and ϵr: C=4π8.8510128310261023102=4π8.851010=8.91109F So, the capacitance in the case when ϵr=8 is C=8.91109F. #b. Find C when ϵr=1.0 for 0<ϕ<π/2 and ϵr=8 for π/2<ϕ<2π.#
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 ϕ. We need to divide the space in two regions with ϵr1=1 (0<ϕ<π/2) and ϵr2=8 (π/2<ϕ<2π). To find the total potential difference, we have to sum the potential differences for both regions: ΔV=ΔV1+ΔV2
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: ΔV1=Q4πϵ0ϵr1[1a1b] ΔV2=Q4πϵ0ϵr2[1a1b] Sum these two to get the total potential difference: ΔV=Q4πϵ0(1ϵr1+1ϵr2)[1a1b]
06

Calculate the capacitance.

Use the formula C=QΔV again: C=QQ4πϵ0(1ϵr1+1ϵr2)(1a1b)=4πϵ0ϵr1ϵr2ϵr1+ϵr2abba Plug the values of a, b, ϵ0, ϵr1, and ϵr2: C=4π8.85101281+8310261023102=4π7.0751010=8.87109F So, the capacitance in the case when ϵr=1.0 for 0<ϕ<π/2 and ϵr=8 for π/2<ϕ<2π is C=8.87109F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (ϵr).
  • 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 ϵr=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=Q4πϵ0ϵrr2
  • 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/r2), 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 ϵr, 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 ΔV is given by:ΔV=Q4πϵ0ϵr[1a1b]
  • This shows how potential difference depends on the dielectric material (ϵr) 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.

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