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Chapter 4: Problem 13

(a) Find the charge density \(\rho(x, y, z)\) that produces the electric field $$ \mathbf{E}=g(\mathbf{i} x+\mathbf{j} \mathbf{y}+\mathbf{k} z) $$ where \(g\) is a constant. (b) Find an electrostatic potential \(\Phi\) such that \(-\nabla \Phi\) is the field \(\mathbf{E}\) given in (a). (c) Verify that \(\nabla^{2} \Phi=-\rho / \epsilon_{0}\)

Short Answer

Expert verified
(a) \( \rho = 3g \epsilon_0 \). (b) \( \Phi = -\frac{1}{2}g(x^2 + y^2 + z^2) + C \). (c) Verified.

Step by step solution

01

Calculate Divergence of E

The electric field is given as \( \mathbf{E} = g(\mathbf{i} x + \mathbf{j} y + \mathbf{k} z) \). To find the charge density \( \rho(x, y, z) \), we need to calculate the divergence of \( \mathbf{E} \). The divergence of a vector field is calculated as \( abla \cdot \mathbf{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \).\- Here, \( E_x = gx \), \( E_y = gy \), and \( E_z = gz \).\- Thus, \( abla \cdot \mathbf{E} = \frac{\partial (gx)}{\partial x} + \frac{\partial (gy)}{\partial y} + \frac{\partial (gz)}{\partial z} = g + g + g = 3g \).
02

Use Gauss's Law to Find Charge Density

According to Gauss's law in differential form, \( abla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \).\- We have calculated \( abla \cdot \mathbf{E} = 3g \).\- Therefore, \( \frac{\rho}{\epsilon_0} = 3g \).\- Solving for \( \rho \), we get \( \rho = 3g \epsilon_0 \).\- Hence, the charge density \( \rho \) is \( 3g \epsilon_0 \).
03

Find Electrostatic Potential \( \Phi \)

The electric field \( \mathbf{E} \) is related to the potential \( \Phi \) by \( \mathbf{E} = -abla \Phi \).\- Given \( \mathbf{E} = g(\mathbf{i} x + \mathbf{j} y + \mathbf{k} z) \), we can find \( \Phi \) by reversing this relationship: \[ abla \Phi = -\mathbf{E} = -g(\mathbf{i} x + \mathbf{j} y + \mathbf{k} z). \] \- Integrate each component: - \( \int (-gx) \, dx = -\frac{1}{2}gx^2 + f(y,z) \) - \( \int (-gy) \, dy = -\frac{1}{2}gy^2 + f(x,z) \) - \( \int (-gz) \, dz = -\frac{1}{2}gz^2 + f(x,y) \)- The potential \( \Phi \) is \( \Phi = -\frac{1}{2}g(x^2 + y^2 + z^2) + C \) where \( C \) is a constant.
04

Verify Poisson's Equation

Poisson's equation states \[ abla^2 \Phi = -\frac{\rho}{\epsilon_0}. \] \- We have \( \Phi = -\frac{1}{2}g(x^2 + y^2 + z^2) + C \). Calculate the Laplacian: - \( abla^2 \Phi = \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} + \frac{\partial^2 \Phi}{\partial z^2} \) - \( \frac{\partial^2}{\partial x^2}(-\frac{1}{2}gx^2) = -\frac{1}{2}g \times 2 = -g \) - Similarly, for the \( y \) and \( z \) components: \( \frac{\partial^2}{\partial y^2}(-\frac{1}{2}gy^2) = -g \) and \( \frac{\partial^2}{\partial z^2}(-\frac{1}{2}gz^2) = -g \)- Thus, \( abla^2 \Phi = -g - g - g = -3g \).- Poisson's equation gives \( -3g = -\frac{\rho}{\epsilon_0} \). Given \( \rho = 3g \epsilon_0 \), this verifies \( abla^2 \Phi = -\rho / \epsilon_0 \).

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Key Concepts

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

Divergence
When dealing with vector fields like electric fields, understanding the concept of divergence is crucial. Divergence essentially measures how much a field spreads from a given point. Think of it like water flowing from a source.
The mathematical operation used is called the 'divergence of a vector field', represented as \( abla \cdot \mathbf{E} \).
For an electric field \( \mathbf{E} = g (\mathbf{i} x + \mathbf{j} y + \mathbf{k} z) \), the divergence can be calculated using the formula:
  • \( abla \cdot \mathbf{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \)
Here, each component of the field contributes, and for a symmetric field like this one, we find that the divergence effectively sums the contributions along all three axes.
This results in the divergence being a constant \( 3g \) for this exercise.
This constant divergence suggests that if we have an infinite source of electric field, it is evenly distributed in space.
Electric Field
An electric field is a vector field around any electric charge, explaining how forces act at a distance. In simple terms, the electric field tells us the direction and magnitude of the force that would act on a positive test charge placed in the field.
The electric field in the given problem is described as \( \mathbf{E} = g (\mathbf{i} x + \mathbf{j} y + \mathbf{k} z) \). This equation shows us that each component (x, y, z) contributes equally to the electric field, scaled by the constant \( g \).
  • Each 'i', 'j', and 'k' signifies the direction along the respective x, y, and z axes.
  • The electric field strength increases linearly with x, y, and z.
The proportional relation indicates a radial field emanating out from the origin, as the field strength increases for points further from the origin. This symmetric spread is key to solving many problems in electrostatics, aligning well with symmetrical charge distributions in space.
Electrostatic Potential
Electrostatic potential, often labeled as \( \Phi \), provides a scalar measure of the potential energy per unit charge due to an electric field.
It simplifies working with electric fields because potentials are easier to add than vector fields.
In this exercise, the relationship between the electric field \( \mathbf{E} \) and electrostatic potential \( \Phi \) is expressed as \( \mathbf{E} = -abla \Phi \).
  • This provides a pathway to find \( \Phi \) when \( \mathbf{E} \) is known.
  • We determine \( \Phi \) by integrating the negative of the field components over their respective variables.
This integration yields the potential function: \( \Phi = -\frac{1}{2}g(x^2 + y^2 + z^2) + C \).
Here, \( C \) is an integration constant that accounts for arbitrary reference potential levels.
This quadratic relationship shows how the potential grows as we move away from the origin, consistent with radial fields, highlighting potential energy increases with distance in such a setup.

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Most popular questions from this chapter

The unit vector normal to the surface \(z=f(x, y)\) is given by $$ \hat{\mathbf{n}}=\left(-\mathbf{i} \frac{\partial f}{\partial x}-\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k}\right) \mid \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}} $$ [see Equation (II-4)]. We have also established that \(\nabla F\) is a vector normal to the surface \(F(x, y, z)=\) const. (page 140 ) so that \(\nabla F /|\nabla F|\) is a unit vector normal to the surface \(F(x, y, z)=\) const. Show that these two expressions for the unit normal vector are identical if \(F(x, y, z)=\) const. and \(z=f(x, y)\) describe the same surface.

Show that \(\nabla \times \nabla f=0\) where \(f(x, y, z)\) is an arbitrary differentiable scalar function. Assume that mixed second-order partial derivatives are independent of the order of differentiation. For example, \(\partial^{2} f / \partial x \partial z\) \(=\partial^{2} f / \partial z \partial x\)

Here is a "proof" that there is no such thing as magnetism. One of Maxwell's equations tells us that $$ \nabla \cdot \mathbf{B}=0 $$ where \(\mathbf{B}\) is any magnetic field. Then using the divergence theorem, we find $$ \iint_{S} \mathbf{B} \cdot \hat{\mathbf{n}} d S=\iiint_{V} \nabla \cdot \mathbf{B} d V=0 $$ Because B has zero divergence, we know (see Problem III-24) there exists a vector function, call it \(\mathbf{A}\), such that $$ \mathbf{B}=\nabla \times \mathbf{A} $$ Combining these last two equations, we get $$ \iint_{s} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Next we apply Stokes' theorem and the preceding result to find $$ \oint_{C} \mathbf{A} \cdot \hat{\mathbf{t}} d s=\iint_{S} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Thus we have shown that the circulation of \(\mathbf{A}\) is path independent. It follows that we can write \(\mathbf{A}=\nabla \psi\), where \(\psi\) is some scalar function. Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that $$ \mathbf{B}=\nabla \times \nabla \psi=0 $$ that is, all magnetic fields are zero! Where did we go wrong? [Taken from G. Arfken, Amer. J. Phys., 27, 526 (1959).]

The function \(\mathbf{F}(r, \theta, z)\) defined in Problem III-17 is smooth and has zero curl in a nonsimply connected region consisting of all of threedimensional space with the \(z\)-axis removed. Show that there is no scalar function \(\psi\) such that \(\mathbf{F}=\nabla \psi\) by evaluating the line integral of \(\mathbf{F} \cdot \hat{\mathbf{t}}\) from the point \(P_{1}(0,-1,0)\) to the point \(P_{2}(0,1,0)\) over two different paths: \(C_{R}\), the right-hand side of the circle of radius 1 lying in the \(x y\)-plane and centered at the origin (see figure), and \(C_{L}\), the lefthand side of the same circle. Orient the paths as shown. Why does the fact that the two paths give different results imply that there is no scalar function \(\psi\) such that \(\mathbf{F}=\nabla \psi ?\)

You are at a point \((a, b, c)\) on the surface $$ z=\left(r^{2}-x^{2}-y^{2}\right)^{1 / 2} \quad(z \geq 0) $$ Assuming both \(a\) and \(b\) are positive, in what direction must you move (a) so that the rate of change of \(z\) will be zero? (b) so that the rate of increase of \(z\) will be greatest? (c) so that the rate of decrease of \(z\) will be greatest? Draw a sketch to show the geometric significance of your answers.
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