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A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 \(\mathrm{cm}\) and the capacitance is 116 \(\mathrm{pF}\) (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is \(220 \mathrm{V},\) what is the magnitude of charge on each sphere?

Short Answer

Expert verified
a) Outer sphere radius is 17.3 cm; b) Charge magnitude is 25.52 nC.

Step by step solution

01

Understand the Formula for Capacitance of a Spherical Capacitor

The capacitance of a spherical capacitor composed of two concentric conducting spheres is given by the formula: \[ C = \frac{4\pi\epsilon_0 R_1 R_2}{R_2 - R_1} \]where \(C\) is the capacitance, \(R_1\) is the radius of the inner sphere, \(R_2\) is the radius of the outer sphere, and \(\epsilon_0 = 8.854 \times 10^{-12} \, \mathrm{F/m}\) is the vacuum permittivity.
02

Solve for Radius of the Outer Sphere

We need to find \(R_2\). First, we rearrange the capacitance formula to solve for \(R_2\):\[ R_2 = \frac{C \cdot R_1}{C - 4\pi\epsilon_0 R_1} \]Substitute in the given values: \(C = 116 \, \mathrm{pF} = 116 \times 10^{-12} \, \mathrm{F}\) and \(R_1 = 0.15 \, \mathrm{m}\):\[ R_2 = \frac{116 \times 10^{-12} \, \mathrm{F} \cdot 0.15 \, \mathrm{m}}{116 \times 10^{-12} \, \mathrm{F} - 4\pi \times 8.854 \times 10^{-12} \, \mathrm{F/m} \cdot 0.15 \, \mathrm{m}} \]Calculate \(R_2\) to find the outer radius.
03

Calculate the Capacitance Using Given Values

Plug the values into the formula from Step 2 to perform the calculation:\[ R_2 = \frac{116 \times 10^{-12} \times 0.15}{116 \times 10^{-12} - 4\pi \times 8.854 \times 10^{-12} \times 0.15} \approx 0.173 \mathrm{m} \]Thus, the radius of the outer sphere is approximately 17.3 cm.
04

Determine the Magnitude of Charge

Using the formula for capacitance \(C = \frac{Q}{V}\), where \(Q\) is the charge and \(V\) is the potential difference, solve for \(Q\):\[ Q = C \cdot V \]Substitute \(C = 116 \times 10^{-12} \, \mathrm{F}\) and \(V = 220 \, \mathrm{V}\):\[ Q = 116 \times 10^{-12} \, \mathrm{F} \times 220 \, \mathrm{V} = 25.52 \times 10^{-9} \, \mathrm{C} \]Thus, the magnitude of the charge on each sphere is approximately \(25.52 \, \mathrm{nC}\).

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Key Concepts

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

Capacitance
Capacitance is a fundamental concept when dealing with capacitors, including our spherical capacitor. It represents the ability of the capacitor to store an electric charge. Imagine a container holding water; similarly, a capacitor stores electric charges. The capacitance is defined by the formula \[ C = \frac{Q}{V} \]where \( C \) is the capacitance, \( Q \) is the charge stored, and \( V \) is the potential difference across the capacitor's terminals.
In the context of the spherical capacitor, the capacitance depends on the radii of the inner and outer spheres and the material (or lack thereof, as in a vacuum). It's also influenced by the vacuum permittivity \( \epsilon_0 \). By understanding and calculating capacitance, we gain insight into how much charge a capacitor can hold per unit of voltage applied.
Potential Difference
The potential difference across a capacitor is essentially the voltage required to store a certain amount of charge. In simpler terms, it's the difference in electric potential (voltage) between two points—in our case, the inner and outer spheres of our spherical capacitor.
Calculating the potential difference is crucial since it helps determine how much charge the capacitor can hold, using the equation:
  • \( Q = C \times V \)
Here, \( V \) represents the potential difference. By adjusting this value, one can control the energy stored in a capacitor, thus making potential difference a vital concept when discussing capacitors and their capabilities.
Vacuum Permittivity
Vacuum permittivity, denoted by \( \epsilon_0 \), is a constant that characterizes how an electric field interacts in a vacuum. It's fundamental to calculating capacitance, particularly in our spherical capacitor problem.
Vacuum permittivity has a value of approximately \( 8.854 \times 10^{-12} \, \mathrm{F/m} \). In the formula for capacitance of spherical capacitors, \( \epsilon_0 \) plays a critical role in determining how well the capacitor can store charge: \[ C = \frac{4\pi\epsilon_0 R_1 R_2}{R_2 - R_1} \]Understanding the concept of vacuum permittivity helps clarify how capacitors behave in different environments, especially when they're separated by vacuum, providing a baseline for electric interactions without the interference of additional materials.
Concentric Spheres
Concentric spheres refer to two spheres that share the same center but have different radii. In our spherical capacitor problem, we have two concentric spherical conducting shells.
The inner sphere is typically smaller, while the outer sphere encompasses the first. These configurations are vital for understanding the physical setup of a spherical capacitor. The capacitance of these concentric spheres depends significantly on their respective radii. Hence, the formula:\[ C = \frac{4\pi\epsilon_0 R_1 R_2}{R_2 - R_1} \]reflects this setup, showing how the difference in radii influences the capacitance. By studying this arrangement, students can grasp how a spherical capacitor's design directly impacts its ability to store electrical energy.

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Most popular questions from this chapter

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{677}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

A budding electronics hobbyist wants to make a simple 1.0 -nf capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of \(3.0,\) and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(12.0 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of poster board? Explain.

A 12.5 -\muF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 \(\mathrm{V}\) . (b) Use Gauss's law (Eq. 24.23 ) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

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