/*! 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 28 CALC In a certain region of spac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

CALC In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y\), where \(A, B\), and \(C\) are positive constants. (a) Calculate the \(x\)-, \(y\)-, and \(z\)-components of the electric field. (b) At which points is the electric field equal to zero?

Short Answer

Expert verified
The electric field is zero at \((-\frac{C}{A}, -\frac{2BC}{A^2}, z)\).

Step by step solution

01

Understanding the relation between electric potential and electric field

The electric field \(\mathbf{E}\) is related to the electric potential \(V(x, y, z)\) by the negative gradient of the potential. Mathematically, this can be expressed as \[\mathbf{E} = -abla V(x,y,z)\] where \(abla V(x, y, z) = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\).
02

Calculating the x-component of the electric field

The x-component of the electric field, \(E_x\), is obtained by computing the partial derivative of the potential V with respect to x: \[E_x = -\frac{\partial V}{\partial x} = -\left(Ay - 2Bx\right) = 2Bx - Ay.\]
03

Calculating the y-component of the electric field

The y-component of the electric field, \(E_y\), is found by computing the partial derivative of the potential V with respect to y: \[E_y = -\frac{\partial V}{\partial y} = -\left(Ax + C\right) = -(Ax + C).\]
04

Calculating the z-component of the electric field

The z-component of the electric field, \(E_z\), is obtained by computing the partial derivative of the potential V with respect to z: \[E_z = -\frac{\partial V}{\partial z} = -0 = 0,\] since \(V(x, y, z)\) has no dependence on z.
05

Finding the points where the electric field is zero

For the electric field \(\mathbf{E} = (E_x, E_y, E_z)\) to be zero, all components must equal zero simultaneously. Therefore, we set:\[2Bx - Ay = 0\]\[-(Ax + C) = 0\]\[E_z = 0\text{ (which is always true since } E_z = 0\text{)}\].Solving for x and y:1. From \(2Bx - Ay = 0\), we have \(2Bx = Ay\) or \(y = \frac{2Bx}{A}\).2. From \(-(Ax + C) = 0\), we have \(Ax + C = 0\) implying \(x = -\frac{C}{A}\).Substituting \(x = -\frac{C}{A}\) into \(y = \frac{2Bx}{A}\) gives \(y = -\frac{2BC}{A^2}\). Therefore, the field is zero at the point \(\left(-\frac{C}{A}, -\frac{2BC}{A^2}, z\right)\).

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.

Electric Field
The electric field is a fundamental concept in electromagnetism. It describes the force that would act on a charged particle at any given point in space. In terms of electric potential, the electric field is mathematically derived as the negative gradient of the potential function. This relationship highlights that an electric field points in the direction in which the potential decreases most rapidly. Therefore, an electric field is not only about the potential in one spot but also how it changes over space. Understanding this component helps unravel how forces operate invisibly through space over charged objects.
Partial Derivatives
Partial derivatives are a vital mathematical tool used to deal with functions of several variables. In our case, we have a potential \(V(x, y, z)\) which is a function of three variables. To find how this potential changes with respect to each of these variables individually, we employ partial derivatives.
  • The partial derivative with respect to \(x\) (denoted as \(\frac{\partial V}{\partial x}\)) considers changes in \(x\) while keeping \(y\) and \(z\) constant.
  • The same concept applies when finding partial derivatives with respect to \(y\) and \(z\).
In the realm of physics, these partial derivatives provide the components of more complex multivariable fields, such as the electric field components we are interested in.
Gradient of Potential
In vector calculus, the gradient of a scalar field is a vector pointing in the direction of the steepest ascent of the field, with a magnitude equal to the rate of increase in that direction. When discussing electric potential \(V\), the gradient is represented as a vector made up of the partial derivatives with respect to each spatial variable: \(abla V = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\). This gradient gives us direct access to the electric field \(\mathbf{E}\) by taking its negative: \(\mathbf{E} = -abla V\).
  • The x-component of the field \(E_x\) relates to \(-\frac{\partial V}{\partial x}\).
  • Similarly, \(E_y\) and \(E_z\) correspond to the other partial derivatives.
By understanding this concept, we can see how variations in potential translate into electric forces in space.
Zero Electric Field Points
The electric field can be zero not only at a single location but also at multiple points in space. For this to occur in our example, each component of the electric field must simultaneously equate to zero. As calculated, the conditions for these zero points involve solving a system of equations derived from setting the partial derivatives to zero.
  • First, solve \(2Bx - Ay = 0\) to establish a relationship between \(x\) and \(y\).
  • Next, use \(-Ax - C = 0\) to determine a specific \(x\) value.
The calculations show that the electric field is zero at the point \(\left(-\frac{C}{A}, -\frac{2BC}{A^2}, z\right)\), indicating specific locations where potential gradients balance out, leaving no force acting on a test charge.

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

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10} \mathrm{~m}\) (a typical atomic distance) to \(3.00 \times 10^{-15} \mathrm{~m}\) (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

Two charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two charges having opposite signs.

A conducting sphere of radius \(R\) has two spherical cavities of radius \(a\) and \(b\). The cavities have charges \(q_{a}\) and \(q_{b}\) respectively at their centres. Find : distance \(r\) (a) The electric field and electric potential at a (i) \(r\) (distance from \(O\), the centre of sphere \(>R\) ) (ii) \(r\) (distance from \(B\), the centre of cavity \(b)

Charges \(+q,-q,+q,-q\) areplaced at corners \(A B C D\) of a square of a side \(a\). A charge \(+q\) is placed at its centre. Find the interaction energy of the system

See all solutions

Recommended explanations on Physics 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.