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Chapter 23: Problem 6

A student argues that the total energy associated with the electric field of a charged sphere must be infinite because its field extends throughout an infinite volume. Critique this argument.

Short Answer

Expert verified
The student's argument is not correct. The total energy associated with the electric field of a charged sphere isn't infinite even though the field extends throughout an infinite volume. This is because the energy isn't uniformly distributed throughout the field. The contribution to total energy from regions far from the charged sphere is very small because the field strength decreases with the square of the distance from the sphere. Therefore, even though the field extends to infinity, the energy does not.

Step by step solution

01

Understanding the concept of electric field energy density

The energy associated with an electric field isn't infinite even though the field might extend throughout an infinite volume. This is because the energy isn't uniformly distributed throughout the field. The intensity of the field decreases with the square of the distance from the source charge. This concept is reflected in the formula for the energy density of an electric field, which is \(U = \frac{1}{2}\epsilon_0 E^2\), where \(U\) is the energy density, \(\epsilon_0\) is the permittivity of free space, and \(E\) is the electric field intensity. The pointer here is that energy density reduces with the square of distance from the source charge because the electric field intensity decreases in that manner.
02

Demonstrating the non-infinite nature of field energy

Taking a uniform sphere charged with charge \(Q\) and radius \(R\), the energy stored in the electric field is given by the formula \(U = \frac{1}{2} \frac{Q^2}{4\pi\epsilon_0 R}\). As can be seen in this equation, though the field exists in all the space outside the sphere, the energy stored in the field doesn't go to infinity. This is because the contribution to the total energy from parts of the field far from the sphere becomes increasingly small due to the \(1/r^2\) dependence of the field strength.

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Most popular questions from this chapter

Your company is still stuck with those 2 - \(\mu\) F capacitors from Problem 50. They turn out to be so cheap that their capacitances are all too low, ranging from \(1.7 \mu \mathrm{F}\) to \(1.9 \mu \mathrm{F}\). A colleague suggests you put variable "trimmer" capacitors in parallel with the cheap capacitors and adjust the combination to precisely \(2.00 \mu \mathrm{F} .\) The available trimmers have variable capacitance from \(25 \mathrm{nF}\) to \(350 \mathrm{nF}\). Will they work?

A parallel-plate capacitor with 1.1 -mm plate spacing has \(\pm 2.3 \mu \mathrm{C}\) on its plates when charged to 150 V. What's the plate area?

A classical view of the electron pictures it as a purely electric entity, whose Einstein rest mass energy, \(E=m c^{2},\) is the energy stored in its electric field. If the electron were a sphere with charge distributed uniformly over its surface, what radius would it have in order to satisfy this condition? (Note: Your answer, and the picture of the electron as a sphere, aren't consistent with quantum theory.)

Two positive point charges are infinitely far apart. Is it possible, using a finite amount of work, to move them until they're a small distance \(d\) apart?

Two widely separated 4.0 -mm-diameter water drops each carry 15 nC. Assuming all charge resides on the drops' surfaces, find the change in electrostatic potential energy if they're brought together to form a single spherical drop.
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