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A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?

Short Answer

Expert verified
Answer: The fractional difference in the capacitances of the spherical capacitor and the parallel plate capacitor is given by \(\frac{r_2 - r_1}{r_1}\), where \(r_1\) and \(r_2\) are the radii of the inner and outer spheres of the spherical capacitor, respectively.

Step by step solution

01

Capacitance of a Spherical Capacitor

To calculate the capacitance of a spherical capacitor made from two thin concentric conducting shells with radii \(r_{1}\) and \(r_{2}\), we use the following formula: \(C_\text{spherical} = 4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1}\) where \(\varepsilon_0\) is the vacuum permittivity.
02

Capacitance of a Parallel Plate Capacitor

To calculate the capacitance of a parallel plate capacitor with the same area as the inner sphere and the same separation \(d = r_2 - r_1\) between the plates, we use the following formula: \(C_\text{parallel} = \frac{\varepsilon_0 A}{d}\) where \(A\) is the area of the plates which is equal to the surface area of the inner sphere: \(A = 4 \pi r_1^2\) Therefore, the capacitance of the parallel plate capacitor is: \(C_\text{parallel} = \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}\)
03

Find the Difference in Capacitances

We will now find the difference between the capacitances of the spherical and parallel plate capacitors: \(\Delta C = C_\text{spherical} - C_\text{parallel}\) Substituting the values, we get: \(\Delta C = 4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1} - \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}\)
04

Calculate the Fractional Difference

Finally, we will calculate the fractional difference between the capacitances. The fractional difference is given by: \(\text{Fractional Difference} = \frac{\Delta C}{C_\text{parallel}}\) Substituting the values, we get: \(\text{Fractional Difference} = \frac{4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1} - \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}}{\frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}}\) Simplifying, we get: \(\text{Fractional Difference} = \frac{r_2 - r_1}{r_1}\) This is the fractional difference in the capacitances of the spherical capacitor and the parallel plate capacitor.

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Key Concepts

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

Spherical Capacitor
A spherical capacitor consists of two concentric spherical conducting shells. These shells have different radii, with the inner shell sized at radius \( r_1 \) and the outer shell at radius \( r_2 \). The unique geometry of a spherical capacitor allows it to store electric charge between the two shells. Since the shells are concentric, the electric field is radial, and this influences how we calculate its capacitance.
To find the capacitance of a spherical capacitor, we utilize the formula:
  • \( C_\text{spherical} = 4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1} \)
Here, \( \varepsilon_0 \) is the vacuum permittivity, a constant that quantifies the ability of the vacuum to permit electric field lines. This formula illustrates how the capacitance depends significantly on the product and difference of the radii of the inner and outer shells. Moreover, the larger the inner radius or the smaller the gap between shells, the greater the capacitance becomes.
Parallel Plate Capacitor
A parallel plate capacitor differs from a spherical capacitor by its geometry. It typically comprises two large, flat conducting plates positioned parallel to one another with a separation \( d \). In our specific problem, the parallel plate capacitor is set up to share the same surface area as the inner sphere and the same gap \( d = r_2 - r_1 \) as the spherical capacitor.
The capacitance of a parallel plate capacitor can be calculated by the formula:
  • \( C_\text{parallel} = \frac{\varepsilon_0 A}{d} \)
Here, \( A \) represents the area of the plates. For this case, the area corresponds to the surface area of the inner sphere, \( A = 4 \pi r_1^2 \). Therefore, the special relationship between plate area and separation allows the configuration to store charge easily if \( r_1 \) is large and \( d \) is small.
Fractional Difference
The concept of fractional difference is significant when comparing two similar entities but of different configurations, like the spherical and parallel plate capacitors. It is a measure detailing how much the capacitance of one system deviates as a fraction of the other. For our capacitors, the fractional difference is defined as:
  • Fractional Difference \( = \frac{C_\text{spherical} - C_\text{parallel}}{C_\text{parallel}} \)
By plugging in the expressions for \( C_\text{spherical} \) and \( C_\text{parallel} \), and simplifying, the fractional difference formula becomes:
  • \( \frac{r_2 - r_1}{r_1} \)
This result indicates how the capacitance of the spherical capacitor varies relative to the parallel plate capacitor depending on the dimensions of the radii \( r_1 \) and \( r_2 \). It highlights when one might be more significantly affected by changes in geometric configuration, pointing towards efficiency in different applications.

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Most popular questions from this chapter

Must a capacitor's plates be made of conducting material? What would happen if two insulating plates were used instead of conducting plates?

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed from between the plates, the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

A parallel plate capacitor is charged with a battery and then disconnected from the battery, leaving a certain amount of energy stored in the capacitor. The separation between the plates is then increased. What happens to the energy stored in the capacitor? Discuss your answer in terms of energy conservation.

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