Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 28

Show that in a homogeneous medium of conductivity \(\sigma\), the potential field \(V\) satisfies Laplace's equation if any volume charge density present does not vary with time.

Short Answer

Expert verified
Answer: In a homogeneous medium with conductivity σ, the potential field V satisfies Laplace's equation if any volume charge density present does not vary with time.

Step by step solution

01

Write down the equation connecting current density to the electric field

The current density \(\vec{J}\) is related to the electric field \(\vec{E}\) by the equation: $$ \vec{J} = \sigma \vec{E} $$ where \(\sigma\) is the conductivity and \(\vec{E}\) is the electric field.
02

Express the electric field in terms of the potential field

The electric field \(\vec{E}\) can be expressed as the negative gradient of the potential field \(V\): $$ \vec{E} = - \nabla V $$
03

Substitute the expression for the electric field into the equation connecting current density to the electric field

Replace \(\vec{E}\) with the expression \(- \nabla V\) to get an equation for the current density in terms of the potential field \(V\): $$ \vec{J} = - \sigma \nabla V $$
04

Express the divergence of the current density in terms of the charge density

According to the charge conservation principle, the divergence of the current density is related to the time derivative of the volume charge density \(\rho\): $$ \nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t} $$
05

Take the divergence of both sides of the equation obtained in Step 3

Apply the divergence operator \(\nabla \cdot\) to both sides of the equation: $$ \nabla \cdot \vec{J} = - \sigma \nabla \cdot (\nabla V) $$
06

Use the charge conservation equation to express the left-hand side as a time derivative of charge density

Replace the term on the left-hand side with the expression given by the charge conservation equation: $$ -\frac{\partial \rho}{\partial t} = - \sigma \nabla \cdot (\nabla V) $$
07

Simplify the equation for a time-invariant charge density

Since the volume charge density does not vary with time, the time derivative of the charge density is zero: $$ 0 = -\sigma \nabla \cdot (\nabla V) $$
08

Divide both sides by the conductivity

Divide both sides of the equation by the conductivity \(\sigma\), which is nonzero in a homogeneous medium: $$ 0 = \nabla \cdot (\nabla V) $$
09

Recognize Laplace's equation

The equation obtained in Step 8 is Laplace's equation for the potential field \(V\): $$ \nabla^2 V = 0 $$ In a homogeneous medium of conductivity \(\sigma\), the potential field \(V\) satisfies Laplace's equation if any volume charge density present does not vary with time.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Coaxial conducting cylinders are located at \(\rho=0.5 \mathrm{~cm}\) and \(\rho=1.2 \mathrm{~cm}\). The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at \(100 \mathrm{~V}\) and the outer at \(0 \mathrm{~V}\), find (a) the location of the \(20 \mathrm{~V}\) equipotential surface; \((b) E_{\rho \max } ;(c) \epsilon_{r}\) if the charge per meter length on the inner cylinder is \(20 \mathrm{nC} / \mathrm{m}\).

Consider an arrangement of two isolated conducting surfaces of any shape that form a capacitor. Use the definitions of capacitance (Eq. (2) in this chapter) and resistance (Eq. (14) in Chapter 5) to show that when the region between the conductors is filled with either conductive material (conductivity \(\sigma\) ) or a perfect dielectric (permittivity \(\epsilon\) ), the resulting resistance and capacitance of the structures are related through the simple formula \(R C=\epsilon / \sigma .\) What basic properties must be true about both the dielectric and the conducting medium for this condition to hold for certain?

A potential field in free space is given as \(V=100 \ln \tan (\theta / 2)+50 \mathrm{~V}\). \((a)\) Find the maximum value of \(\left|\mathbf{E}_{\theta}\right|\) on the surface \(\theta=40^{\circ}\) for \(0.1

(a) Determine the capacitance of an isolated conducting sphere of radius \(a\) in free space (consider an outer conductor existing at \(r \rightarrow \infty\) ). \((b)\) The sphere is to be covered with a dielectric layer of thickness \(d\) and dielectric contant \(\epsilon_{r}\). If \(\epsilon_{r}=3\), find \(d\) in terms of \(a\) such that the capacitance is twice that of \(\operatorname{part}(a) .\)

Let \(V=(\cos 2 \phi) / \rho\) in free space. (a) Find the volume charge density at point \(A\left(0.5,60^{\circ}, 1\right) .(b)\) Find the surface charge density on a conductor surface passing through the point \(B\left(2,30^{\circ}, 1\right)\).
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Astrophysics

Read Explanation

Kinematics Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free