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24.68. A solid conducting sphere of radius \(R\) carries a charge \(Q\) . Calculate the electric-field energy density at a point a distance \(r\) from the center of the sphere for (a) \(rR\) (c) Calculate the total electric-field energy associated with the charged sphere. (Hint: Consider a spherical shell of radius \(r\) and thickness \(d r\) that has volume \(d V=4 \pi r^{2} d r,\) and find the energy stored in this volume. Then integrate from \(r=0\) to \(r \rightarrow \infty .\) (d) Explain why the result of part (c) can be interpreted as the amount of work required to assemble the charge \(Q\) on the sphere. (e) By using Eq. \((249)\) and the result of part (c), show that the capacitance of the sphere is as given in Problem 24.67 .

Short Answer

Expert verified
The electric field is 0 inside and \( \frac{kQ}{r^2} \) outside the sphere. Total energy is \( \frac{kQ^2}{2R} \), and capacitance is \( 4\pi\varepsilon_0 R \).

Step by step solution

01

Understanding the electric field inside the sphere (r

Inside a solid conducting sphere, the electric field is zero due to the uniform distribution of charge on the surface. Thus, for any point inside the sphere where \( r < R \), the electric-field energy density is \( u = \frac{1}{2}\varepsilon_0 E^2 = 0 \).
02

Calculating electric field outside the sphere (r>R)

For \( r > R \), the electric field due to the charged sphere acts as if the entire charge \( Q \) is concentrated at the center. This gives \( E = \frac{kQ}{r^2} \), where \( k = \frac{1}{4\pi\varepsilon_0} \). Substitute in the electric-field energy density formula: \( u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2}\varepsilon_0 \left( \frac{kQ}{r^2} \right)^2 \).
03

Integrating to find total electric-field energy

To find the total energy, consider a shell of thickness \( dr \) at distance \( r \). The energy in this shell is \( dU = u \cdot dV = \frac{1}{2}\varepsilon_0 \left(\frac{kQ}{r^2}\right)^2 \cdot 4\pi r^2 dr \). Integrating from \( R \) to \( \infty \) yields the total energy \( U = \int_R^\infty \frac{k^2Q^2}{8\pi\varepsilon_0 r^2} \cdot 4\pi r^2 dr = \frac{kQ^2}{2R} \).
04

Interpretation of energy as work done

This total electric-field energy, \( \frac{kQ^2}{2R} \), represents the work done to assemble the charge \( Q \) on the sphere from infinity, due to the energy needed to bring each differential charge \( dq \) to the sphere against the potential created by the charges already present.
05

Using total energy to find capacitance

The capacitance of a sphere is given by \( C = \frac{Q}{V} \), where \( V \) is the potential. From the energy expression, the potential \( V = \frac{kQ}{R} \). So, combining this, we get \( C = 4\pi\varepsilon_0 R \), confirming the problem's statement.

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Key Concepts

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

Electric Field Energy Density
Electric field energy density is the amount of energy stored per unit volume in an electric field. It is a critical concept when dealing with charged objects and understanding how energy is distributed in space around them. The formula to calculate the electric field energy density, denoted as \( u \), is:\[ u = \frac{1}{2} \varepsilon_0 E^2 \]where \( \varepsilon_0 \) is the permittivity of free space, and \( E \) is the electric field strength.
  • Inside a conducting sphere with radius \( R \), the electric field is zero throughout the interior \(( r < R )\). Hence, the energy density inside is also zero.
  • Outside the sphere, where \( r > R \), the electric field behaves as if all the charge \( Q \) is concentrated at the center of the sphere. The energy density can then be expressed as: \( u = \frac{1}{2} \varepsilon_0 \left( \frac{kQ}{r^2} \right)^2\).
This expression indicates how energy is dispersed variably in the space surrounding the charged sphere, declining with the square of the distance from the sphere.
Capacitance of a Sphere
The capacitance of an isolated sphere represents its ability to store electrical charge. It relates the charge \( Q \) on the sphere to the electric potential \( V \). The formula for capacitance \( C \) of a sphere with radius \( R \) is:\[ C = 4\pi\varepsilon_0 R \]To derive this, consider the potential \( V \) of the sphere, caused when charge \( Q \) is distributed across its surface. The potential can be represented as:\[ V = \frac{kQ}{R} \]Using the relationship \( C = \frac{Q}{V} \) and substituting for \( V \), we obtain:
  • \( C = \frac{Q}{\frac{kQ}{R}} = \frac{Q}{\frac{Q}{4\pi\varepsilon_0 R}} \)
  • This simplifies to \( C = 4\pi\varepsilon_0 R \), illustrating that the capacitance depends linearly on the sphere's radius.
Thus, the larger the sphere, the higher its capacitance, resulting in a greater ability to hold charge for a given potential.
Integration of Electric Field
To find the total electric energy stored in the electric field surrounding a conducting sphere, we integrate the energy density over the space outside the sphere. Using a spherical shell of radius \( r \) and thickness \( dr \), the incremental volume \( dV \) is:\[ dV = 4\pi r^2 dr \]The energy stored in this shell, \( dU \), is:\[ dU = u \cdot dV = \frac{1}{2} \varepsilon_0 \left( \frac{kQ}{r^2} \right)^2 \cdot 4\pi r^2 dr \]When integrated from the surface of the sphere \( R \) to infinity, we find the total energy \( U \):\[ U = \int_R^\infty \frac{k^2Q^2}{8\pi\varepsilon_0 r^2} \cdot 4\pi r^2 dr \]This simplifies and evaluates to:\[ U = \frac{kQ^2}{2R} \]This result signifies the total work done to assemble the charge \( Q \) onto the sphere, providing valuable insight into both theoretical and practical contexts of electric field applications. This integration also serves to illuminate how electric fields behave and interact over distance, influencing the distribution of forces and energy in space.

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Most popular questions from this chapter

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (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 \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.8. A \(5.00-\) pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between the plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

24.74. The parallel-plate air capacitor in Fig. 24.40 consists of two horizontal conducting plates of equal area \(A .\) The bottom plate rests on a fixed support, and the top plate is suspended by four springs with spring constant \(k,\) positioned at each of the four corners of the top plate as shown in the figure. When uncharged, the plates are separated by a distance \(z_{0}\) . A battery is connected to the plates and produces a potential difference \(V\) between them. This causes the plate separation to decrease to \(z\) . Neglect any fringing effects. (a) Show that the electrostatic force between the charged plates has a magnitude \(\epsilon_{0} A V^{2} / 2 z^{2}\) . (Hint: See Exercise \(24.29 . )\) (b) Obtain an expression that relates the plate separation \(z\) to the potential difference \(V\) . The resulting equation will be cubic in \(z\) . (c) Given the values \(A=0.300 \mathrm{m}^{2}, z_{0}=1.20 \mathrm{mm}, k=25.0 \mathrm{N} / \mathrm{m},\) and \(V=120 \mathrm{V},\) find the two values of \(z\) for which the top plate will be in equilibrium. (Hint: You can solve the cubic equation by plugging a trial value of \(z\) into the equation and then adjusting your guess until the equation is satisfied to three significant figures. Locating the roots of the cubic equation graphically can help you pick starting values of \(z\) for this trial-and error procedure. One root of the cubic equation has a nonphysical negative value. (d) For each of the two values of \(z\) found in part (c), is the equilibrium stable or unstable? For stable equilibrium a small displacement of the object will give rise to a net force tending to return the object to the equilibrium position. For unstable equilibrium a small displacement gives rise to a net force that takes the object farther away from equilibrium.

24.48 . A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Tefion. (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.

24.27. A \(450-\mu\) F capacitor is charged to 295 V. Then a wire is con- nected between the plates. How many joules of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire?

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