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Chapter 6: Problem 6

Find the energy stored in a sphere of charge \(q\) and radius \(a\) with uniform charge density, and show that infinite energy is required to compress the sphere to a point. Find also the stored energy in the case where the charge is uniformly spread over the surface of the sphere.

Short Answer

Expert verified
Answer: For a sphere with uniform charge density, the energy stored is \(U = \frac{3q^2}{5\epsilon_0 a}\). For a sphere with charge on its surface, the energy stored is \(U = \frac{q^2}{8\pi\epsilon_0 a}\).

Step by step solution

01

Electric Field of a Uniformly Charged Sphere

First, let's calculate electric potential inside (\(r < a\)) and outside (\(r > a\)) the uniformly charged sphere. For a sphere with a charge density \(\rho = \frac{3q}{4\pi a^3}\) and radius \(a\), - Inside: \(V_{in}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{2a^3}(3a^2 - r^2)\) - Outside: \(V_{out}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}\)
02

Calculate the Energy Stored in the Sphere

Now, we can use the expression for the potential energy: \(U = \frac{1}{2} \int_{V} \rho(\vec{r}) V(\vec{r}) d^3\vec{r}\) We separate the integral into two parts (inside and outside the sphere) and consider that \(\rho_{out} = 0\): \(U = \frac{1}{2} \int_{V_{in}} \rho(\vec{r}) V_{in}(\vec{r}) d^3\vec{r} + 0\) Now, substitute the values for \(\rho(\vec{r})\) and \(V_{in}(\vec{r})\): \(U = \frac{1}{2} \int_{V_{in}} \frac{3q}{4\pi a^3} \cdot \frac{1}{4\pi\epsilon_0}\frac{q}{2a^3}(3a^2 - r^2) d^3\vec{r}\) Now integrate: \(U = \frac{3q^2}{40\pi\epsilon_0 a^5} \int_{0}^{a} (3a^2 - r^2) r^2\, 4\pi r^2 dr\) \(U = \frac{3q^2}{10\epsilon_0 a^5} \int_{0}^{a} (3a^2 - r^2) r^4 dr\) After integrating over the limits, we get: \(U = \frac{3q^2}{5\epsilon_0 a}\) This expression shows the energy stored in a uniformly charged sphere.
03

Compressing the Sphere to a Point

To analyze the energy required to compress the sphere to a point, we can take the limit as the sphere's radius approaches zero: \(\lim_{a \to 0} \frac{3q^2}{5\epsilon_0 a}\) This limit is infinite, which means that infinite energy is required to compress the sphere to a point.
04

Energy Stored with Charge on the Surface

Now let's consider the case when the charge is uniformly spread over the surface of the sphere. The surface charge density is given by: \(\sigma = \frac{q}{4\pi a^2}\) The expression for potential energy in this case is: \(U = \frac{1}{2} \int_{S} \sigma(\vec{r}) V(\vec{r}) d^2\vec{r}\) While the sphere has a uniform surface charge distribution, the electric potential is constant across the sphere: \(V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{a}\) Now substitute the values for \(\sigma(\vec{r})\) and \(V(\vec{r})\) and integrate over the sphere's surface: \(U = \frac{1}{2} \int_{S} \frac{q}{4\pi a^2} \cdot \frac{1}{4\pi\epsilon_0}\frac{q}{a} d^2\vec{r}\) \(U = \frac{q^2}{32\pi^2\epsilon_0 a^3} \int_{S} d^2\vec{r}\) After integrating over the surface, we get: \(U = \frac{q^2}{8\pi\epsilon_0 a}\) This expression shows the energy stored in a sphere with uniform charge distribution on its surface.

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Key Concepts

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

Electric Potential
Imagine electric potential as a measure of the work needed to move a charge within an electric field, similar to lifting an object against gravity. The potential at any point around a charged sphere is essentially the energy per unit charge at that point.

In the context of our exercise, the electric potential inside a uniformly charged sphere (\r < a) is expressed as \(V_{in}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{2a^3}(3a^2 - r^2)\), considering a charge density \(\rho\), and outside (\r > a) as \(V_{out}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}\). As we approach the sphere's surface from the inside, the electric potential reaches its maximum, indicating the most work done per unit charge to bring it to that point from infinity.
Uniform Charge Density
Uniform charge density refers to the even distribution of electrical charge throughout a material. In our sphere, this means that at any cut of the sphere, the charge spread will be consistent, giving us the density \(\rho = \frac{3q}{4\pi a^3}\).

Understanding this uniformity is crucial, as it directly affects the electric potential and field calculations. Because of this even spread of charge within its volume, we can integrate the potential over the entire volume to find the total potential energy, without concerning ourselves with varying charge densities.
Electric Field
The electric field describes how the charge disperses its influence through space, affecting other charges around it. It's like the magnetic field around a magnet, invisible to the eye but crucially important for the forces experienced by nearby charges.

In our charged sphere, the field exhibits spherical symmetry and is related to the electric potential via the gradient. For every point inside the sphere, the electric field is pointing outward and changes with the distance from the center, as governed by the charge density. This is what ultimately ties together the concepts of charge density and electric potential, with the field encapsulating both properties.
Potential Energy
Potential energy, in this electric context, is the energy a charge holds due to its position within an electric field. Imagine a ball at the top of a hill鈥攅lectric potential energy is analogous, but instead of gravity, it's the electric field doing the work.

By integrating the product of charge density and electric potential over the volume of the sphere, we calculate the total potential energy stored, represented by \(U = \frac{3q^2}{5\epsilon_0 a}\) for a uniformly charged sphere. When considering a sphere compressed to a point, the notion that infinite energy is required suggests that charges naturally repel each other to avoid such a singularity, an example of the principles governing fundamental particle interactions.

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Most popular questions from this chapter

Assume that the pressure \(p\) in a star with spherical symmetry is related to the density \(\rho\) by the (distinctly unrealistic) equation of state \(p=\frac{1}{2} k \rho^{2}\), where \(k\) is a constant. Use the fluid equilibrium equation obtained in Problem 23 to find a relation between \(\rho\) and \(\Phi .\) Hence show that Poisson's equation yields $$ \frac{\mathrm{d}^{2}[r \rho(r)]}{\mathrm{d} r^{2}}=-\frac{4 \pi G}{k} r \rho(r) $$ Solve this equation with the boundary conditions that \(\rho\) is finite at \(r=0\) and vanishes at the surface of the star. Hence show that the radius \(a\) of the star is determined solely by \(k\) and is independent of its mass \(M\). Show also that \(M=(4 / \pi) a^{3} \rho(0)\).

Assume that the Earth consists of a core of uniform density \(\rho_{\mathrm{c}}\), surrounded by a mantle of uniform density \(\rho_{\mathrm{m}}\), and that the boundary between the two is of similar shape to the outer surface, but with a radius only three-fifths as large. Find what ratio of densities \(\rho_{c} / \rho_{\mathrm{m}}\) is required to explain the observed quadrupole moment. (Hint: Treat the Earth as a superposition of two ellipsoids of densities \(\rho_{\mathrm{m}}\) and \(\rho_{\mathrm{c}}-\rho_{\mathrm{m}}\). Note that in reality neither core nor mantle is of uniform density.)

Show that the moment of the Earth's gravitational force may be written in the form $$ m \boldsymbol{r} \wedge \boldsymbol{g}=\frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$ Consider a satellite in a circular orbit of radius \(r\) in a plane inclined to the equator at an angle \(\alpha\). By introducing a pair of axes in the plane of the orbit, as in \(\S 5.6\) (see Fig. 5.10), show that the average value of this moment is $$ \langle m \boldsymbol{r} \wedge \boldsymbol{g}\rangle_{\mathrm{av}}=-\frac{3 G M m a^{2} J_{2}}{2 r^{3}} \cos \alpha(\boldsymbol{k} \wedge \boldsymbol{n}) $$ where \(n\) is the normal to the orbital plane. Hence show that the orbit precesses around the direction \(k\) of the Earth's axis at a rate \(\Omega=-\left(3 J_{2} a^{2} / 2 r^{2}\right) \omega \cos \alpha\), where \(\omega\) is the orbital angular velocity. Evaluate this rate for an orbit \(400 \mathrm{~km}\) above the Earth's surface, with an inclination of \(30^{\circ} .\) Find also the precessional period.

The potential \(\phi(\boldsymbol{r})=\left(q / 4 \pi \epsilon_{0} r\right) \mathrm{e}^{-\mu r}\) may be regarded as representing the effect of screening of a charge \(q\) at the origin by mobile charges in a plasma. Calculate the charge density \(\rho\) (at points where \(r \neq 0\) ) and find the total charge throughout space, excluding the origin.

The rotation period of Jupiter is approximately 10 hours. Its mass and radius are \(318 M_{\mathrm{E}}\) and \(11.0 R_{\mathrm{E}}\), respectively \((\mathrm{E}=\) Earth). Calculate approximately its oblateness, neglecting the variation in density. (The observed value is about \(1 / 15\).)
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