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Ch 26: Potential and Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 26, Problem 81a

Find an expression for the capacitance of a spherical capacitor, consisting of concentric spherical shells of radii R1 (inner shell) and R2 (outer shell).

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Start by recalling the formula for the capacitance of a capacitor: \( C = \frac{Q}{\Delta V} \), where \( Q \) is the charge on the capacitor and \( \Delta V \) is the potential difference between the two shells.
Determine the electric field \( E \) between the two spherical shells using Gauss's law: \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\varepsilon_0} \). For a spherical shell, the electric field at a distance \( r \) from the center is \( E = \frac{Q}{4 \pi \varepsilon_0 r^2} \).
Find the potential difference \( \Delta V \) between the two shells by integrating the electric field from \( R_1 \) to \( R_2 \): \( \Delta V = \int_{R_1}^{R_2} E \, dr = \int_{R_1}^{R_2} \frac{Q}{4 \pi \varepsilon_0 r^2} \, dr \). Perform the integration to get \( \Delta V = \frac{Q}{4 \pi \varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).
Substitute \( \Delta V \) into the capacitance formula \( C = \frac{Q}{\Delta V} \): \( C = \frac{Q}{\frac{Q}{4 \pi \varepsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \). Simplify the expression to get \( C = 4 \pi \varepsilon_0 \left( \frac{1}{\frac{1}{R_1} - \frac{1}{R_2}} \right) \).
Simplify further to express the capacitance in terms of \( R_1 \) and \( R_2 \): \( C = \frac{4 \pi \varepsilon_0 R_1 R_2}{R_2 - R_1} \). This is the final expression for the capacitance of the spherical capacitor.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined as the ratio of the electric charge (Q) stored on one conductor to the potential difference (V) between the conductors. The unit of capacitance is the farad (F), where 1 farad equals 1 coulomb per volt.
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Electric Field

The electric field (E) is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. For a spherical capacitor, the electric field between the two shells can be derived from Gauss's law, which relates the electric field to the charge enclosed by a Gaussian surface. The electric field is crucial for understanding how the capacitor stores energy.
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Gauss's Law

Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. It is mathematically expressed as ∮E·dA = Q_enc/ε₀, where E is the electric field, dA is the differential area, Q_enc is the enclosed charge, and ε₀ is the permittivity of free space. This law is essential for deriving the electric field in spherical capacitors and ultimately calculating their capacitance.
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