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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 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

Short Answer

Expert verified
(a) The radius of the outer sphere is approximately 16.7 cm. (b) The magnitude of the charge on each sphere is about 25.5 nC.

Step by step solution

01

Understanding Capacitance of a Spherical Capacitor

The capacitance of a spherical capacitor is given by the formula \( C = \frac{4\pi\varepsilon_0\cdot a\cdot b}{b-a} \), where \( a \) is the radius of the inner sphere, \( b \) is the radius of the outer sphere, and \( \varepsilon_0 \) is the permittivity of free space. Here, \( C = 116 \) pF and the radius \( a = 15.0 \) cm.
02

Rearranging the Formula for the Outer Radius

To find the radius of the outer sphere \( b \), rearrange the formula: \( b = \frac{C \cdot a}{C - 4\pi\varepsilon_0 \cdot a} + a \). We need to ensure all units are consistent for calculation.
03

Inserting Values and Solving for Outer Radius

Convert 15.0 cm to meters (0.15 m) and plug the values into the formula from Step 2. With \( C = 116 \times 10^{-12} \) F and \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m, we solve for \( b \).
04

Calculating the Permittivity Substitution

Calculate \( 4\pi\varepsilon_0 \cdot a \) as \( 4\pi \times 8.85 \times 10^{-12} \times 0.15 \). Substitute this back into the equation from Step 3.
05

Solving the Capacitance Equation

Perform the arithmetic to find \( b \, - \, a \), and consequently \( b \). Finally, solve for the unknown \( b \) and convert to cm.
06

Determine the Charge on Each Sphere

Using the formula \( Q = C \cdot V \), find the charge \( Q \) by multiplying the capacitance \( C = 116 \times 10^{-12} \) F by the potential difference \( V = 220 \) V.
07

Solving the Charge Equation

Calculate \( Q = 116 \times 10^{-12} \times 220 \). Solve this to find \( Q \) in Coulombs.

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Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store electric charge. It is denoted by the symbol \( C \) and is usually measured in farads (F). For a capacitor, the capacitance is directly related to its geometry and the permittivity of the medium between the conductors.

In the case of a spherical capacitor, which consists of two concentric spherical conducting shells, the capacitance formula is given by:
\[C = \frac{4\pi\varepsilon_0 \cdot a \cdot b}{b-a}\]
where \( a \) is the radius of the inner sphere, \( b \) is the radius of the outer sphere, and \( \varepsilon_0 \) is the permittivity of free space. In this formula, the difference \( b-a \) represents the separation between the spheres, controlling how electric fields interact. A high capacitance means more charge can be stored at a given potential difference.

Capacitance is also influenced by the material between the plates (or spheres in this case). In our exercise, the space is a vacuum which standardizes calculations using \( \varepsilon_0 \).
Potential Difference
The potential difference, often called voltage, is the work needed to move a unit charge between two points in a field. Denoted by \( V \), it is measured in volts and is a fundamental concept in understanding how capacitors function.

For capacitors, the potential difference is related to the amount of electric charge stored. In mathematical terms, it is described by the formula:
\[V = \frac{Q}{C}\]
where \( Q \) is the electric charge stored in the capacitor and \( C \) is the capacitance. This relation signifies that for a given amount of stored charge, a larger capacitance will lead to a smaller potential difference and vice versa. In our exercise, the potential difference between the spherical shells is significant for calculating the stored charge.

Understanding potential differences is crucial when determining the operation conditions of capacitors in electrical circuits and systems.
Electric Charge
Electric charge, denoted by \( Q \), is a fundamental property of matter that causes it to experience a force when near other electrically charged matter. In capacitors, it refers to the quantity of electrical charge stored on the plates or shells.

The relationship between charge, capacitance, and potential difference is expressed as:
\[Q = C \cdot V\]
This formula shows that the charge \( Q \) stored in a capacitor is a product of the capacitance \( C \) and the potential difference \( V \) across the capacitor.

In our scenario, to find the charge stored in the spherical capacitor, one would multiply its capacitance (116 pF in the problem) by the potential difference (220 V). The calculation tells us how much charge each sphere of the capacitor holds under the given conditions, highlighting the utility of capacitors in managing electrical energy.
Permittivity of Free Space
The permittivity of free space, symbolized as \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum between conductor surfaces. Its value is approximately \( 8.85 \times 10^{-12} \) F/m (farads per meter).

This constant plays a central role in determining the capacitance of capacitors, as seen in the capacitance formula for spherical capacitors:
\[C = \frac{4\pi\varepsilon_0 \cdot a \cdot b}{b-a}\]
It essentially relates the geometric factors (radii of the spheres) to the potential to store charge through an electric field in a vacuum.

Understanding \( \varepsilon_0 \) is vital not only in theoretical physics but also in practical applications involving insulators and capacitors. It helps in estimating the efficiency of capacitors in storing energy and ensures precision in computations involving electric and magnetic fields.

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Most popular questions from this chapter

A parallel-plate capacitor has plates with area 0.0225 m\(^2\) separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 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.

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

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 mm. (a) If the sheets of paper measure 22 \(\times\)28 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 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 posterboard? Explain.

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