/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Potential functions arise freque... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(V=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla V,\) where \(\nabla V\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

Short Answer

Expert verified
Question: Calculate the magnitude of the electric field vector for the given potential function, and explain why it is referred to as an inverse square law. Solution: 1. Calculate the gradient of the potential function V = kQ/r in three dimensions. 2. Negate the gradient to find the electric field vector. 3. Determine the magnitude of the electric field vector. 4. Observe that the magnitude is inversely proportional to the square of the distance r (|E| = kQ/r^2). The term "inverse square law" refers to the fact that the magnitude of the electric field decreases with the square of the distance between the point charge and the point of interest, such that any increase in distance results in rapid decrease in electric field strength.

Step by step solution

01

Calculate the gradient of V

Firstly, we need to find the gradient of the potential function \(V\). The gradient of a function in three dimensions is given by: $$ \nabla V=\left\langle\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right\rangle $$ Recall that the potential function \(V=k Q / r\), and \(r^{2}=x^{2}+y^{2}+z^{2}\). So, we will find the partial derivatives of \(V\) with respect to \(x\), \(y\), and \(z\). $$ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$ $$ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$ $$ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$
02

Evaluate the partial derivatives

Now we'll evaluate the three partial derivatives. Using the chain rule, we have: $$ \frac{\partial V}{\partial x} = -kQ\frac{x}{(x^2+y^2+z^2)^{3/2}} $$ $$ \frac{\partial V}{\partial y} = -kQ\frac{y}{(x^2+y^2+z^2)^{3/2}} $$ $$ \frac{\partial V}{\partial z} = -kQ\frac{z}{(x^2+y^2+z^2)^{3/2}} $$
03

Find the electric field vector

Since \(\mathbf{E}=-\nabla V\), we can now find the electric field vector by plugging in the gradient components: $$ \mathbf{E}(x, y, z) = kQ \left\langle \frac{x}{r^3},\frac{y}{r^3},\frac{z}{r^3} \right\rangle $$ b. Finding the magnitude of the electric field and explaining the inverse square law.
04

Calculate the magnitude of the electric field vector

To find the magnitude of the electric field vector, recall the formula for the magnitude of a vector: $$ |\mathbf{E}| = \sqrt{(E_x)^2+(E_y)^2+(E_z)^2} $$ Plugging in the components of \(\mathbf{E}\), we get: $$ |\mathbf{E}| = \sqrt{\left(kQ\frac{x}{r^3}\right)^2+\left(kQ\frac{y}{r^3}\right)^2+\left(kQ\frac{z}{r^3}\right)^2} $$ Simplifying, we obtain: $$ |\mathbf{E}| = \frac{kQ}{r^2} $$
05

Explain why it is called an inverse square law

The relationship \(|\mathbf{E}| = kQ/r^2\) is called an inverse square law because the magnitude of the electric field \(|\mathbf{E}|\) is inversely proportional to the square of the distance \(r^2\) between the point charge and the point of interest. This means that as the distance away from the point charge doubles, the electric field's strength decreases by a factor of 4, and as the distance increases, the electric field strength decreases even more rapidly.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Gradient
The concept of a gradient is critically important in understanding how potential functions relate to physical fields, such as electric fields. A gradient in mathematics refers to a vector that points in the direction of the greatest rate of increase of a scalar function. In three dimensions, it is represented as:
  • \( abla V = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right\rangle \)

This notation indicates that the gradient is comprised of the partial derivatives of the function with respect to each spatial coordinate. In the context of potential functions, the gradient illustrates how steeply the potential increases or decreases at a given point.
For example, the electric field is derived from the potential function as the negative of its gradient. This means that the electric field vector points in the direction that steeply decreases potential, and the magnitude of this vector indicates the strength of the field. The relationship between the potential function and the electric field vector underscores the power of gradients in translating mathematical descriptions into physical realities.
Electric Field
An electric field is a vector field around a charged object where electric forces occur. It signifies the strength and direction that a charge would move if placed in the field. The electric field \( \mathbf{E} \) due to a point charge is computed as the negative gradient of the potential function \( V \). Typically, the potential, \( V \), for a point charge is expressed as:
  • \( V = \frac{kQ}{r} \)

Where \( Q \) is the charge, \( r \) is the distance from the charge, and \( k \) is a constant.
To find the electric field, calculate \( -abla V \). This results in the following expression for the electric field vector:
  • \( \mathbf{E} = kQ \left\langle \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \right\rangle \)

The components \( \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \) reveal how the electric field varies in each spatial direction. This field pattern shows that electric fields emanate outward in symmetrical patterns when looking at a charge in a vacuum. This formulation suggests that the field's direction depends on the spatial coordinates \( (x, y, z) \). Hence, they directly influence the resulting electric field through their relationship to \( r \), the radial distance.
Inverse Square Law
The inverse square law describes how the magnitude of certain physical quantities diminishes with distance. In the case of the electric field from a point charge, this law tells us how the electric force decreases as we move away from the charge. The magnitude of the electric field \( |\mathbf{E}| \) is given as:
  • \( |\mathbf{E}| = \frac{kQ}{r^2} \)

This equation illustrates the essence of the inverse square law: as the distance \( r \) from the charge increases, the magnitude of \( |\mathbf{E}| \) rapidly decreases, specifically by a factor of \( 1/r^2 \).
This relationship emerges from geometric considerations. When the distance doubles, the area through which the electric field lines spread increases by a factor of four (\( 2^2 \)), causing the electric field's strength to drop to one-quarter.
Understanding the inverse square law is fundamental in physics, not just for electrical phenomena but also for gravitational and illumination applications. Such recognition of broad occurrences helps students grasp how physical laws apply in diverse contexts.

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

Determine whether the following statements are true and give an explanation or counterexample. a. The plane passing through the point (1,1,1) with a normal vector \(\mathbf{n}=\langle 1,2,-3\rangle\) is the same as the plane passing through the point (3,0,1) with a normal vector \(\mathbf{n}=\langle-2,-4,6\rangle\) b. The equations \(x+y-z=1\) and \(-x-y+z=1\) describe the same plane. c. Given a plane \(Q\), there is exactly one plane orthogonal to \(Q\). d. Given a line \(\ell\) and a point \(P_{0}\) not on \(\ell\), there is exactly one plane that contains \(\ell\) and passes through \(P_{0}\) e. Given a plane \(R\) and a point \(P_{0},\) there is exactly one plane that is orthogonal to \(R\) and passes through \(P_{0}\) f. Any two distinct lines in \(\mathbb{R}^{3}\) determine a unique plane. g. If plane \(Q\) is orthogonal to plane \(R\) and plane \(R\) is orthogonal to plane \(S\), then plane \(Q\) is orthogonal to plane \(S\).

Production functions Economists model the output of manufacturing systems using production functions that have many of the same properties as utility functions. The family of Cobb-Douglas production functions has the form \(P=f(K, L)=C K^{a} L^{1-a},\) where K represents capital, L represents labor, and C and a are positive real numbers with \(0

Ideal Gas Law Many gases can be modeled by the Ideal Gas Law, \(P V=n R T,\) which relates the temperature \((T,\) measured in Kelvin (K)), pressure ( \(P\), measured in Pascals (Pa)), and volume ( \(V\), measured in \(\mathrm{m}^{3}\) ) of a gas. Assume that the quantity of gas in question is \(n=1\) mole (mol). The gas constant has a value of \(R=8.3 \mathrm{m}^{3} \cdot \mathrm{Pa} / \mathrm{mol} \cdot \mathrm{K}.\) a. Consider \(T\) to be the dependent variable and plot several level curves (called isotherms) of the temperature surface in the region \(0 \leq P \leq 100,000\) and \(0 \leq V \leq 0.5.\) b. Consider \(P\) to be the dependent variable and plot several level curves (called isobars) of the pressure surface in the region \(0 \leq T \leq 900\) and \(0< V \leq 0.5.\) c. Consider \(V\) to be the dependent variable and plot several level curves of the volume surface in the region \(0 \leq T \leq 900\) and \(0 < P \leq 100,000.\)

Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

See all solutions

Recommended explanations on Math Textbooks

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.