Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 34

Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. Are there any points not on the \(x\) -axis where $\overrightarrow{\mathbf{E}}=0 ?$ Explain.

Short Answer

Expert verified
Based on the above solution, the analysis and equations show that there are no points not on the x-axis where the electric field is equal to zero. This conclusion is drawn from equation (5) which demonstrates that it is independent of θ1 and θ2 and thus does not satisfy the condition of having y≠0.

Step by step solution

01

Electric field due to individual charges

The electric field at a point P due to a single point charge q is given by \(\vec{E} = \dfrac{k*q}{r^2}\hat{r}\), where \(k\) is the electrostatic constant, \(q\) is the charge, \(r\) is the distance between the charge and the point P, and \(\hat{r}\) is the unit vector pointing from the charge to P.
02

Components of electric fields

Let's consider a point P in the x-y plane that is not on the x-axis (i.e., \(y\neq0\)). Let \(r_1\) and \(r_2\) denote the distances from the point P to the charges \(q\) and \(2q\), respectively. We will find the components of the electric fields \(\vec{E}_1\) and \(\vec{E}_2\) due to these two charges along the x and y directions. Using trigonometric functions, we have, For \(\vec{E}_1\): \(E_{1x} = \dfrac{k*q}{r_1^2}\cos{\theta_1}\) and \(E_{1y} = \dfrac{k*q}{r_1^2}\sin{\theta_1}\) For \(\vec{E}_2\): \(E_{2x} = -\dfrac{k*2q}{r_2^2}\cos{\theta_2}\) and \(E_{2y} = \dfrac{k*2q}{r_2^2}\sin{\theta_2}\) The negative sign in \(E_{2x}\) is because its x-component is directed opposite to the \(x\)-axis.
03

Equating the x and y components of the electric fields

To find the point P where the total electric field becomes zero, we first equate both the x-components and then the y-components of the electric fields: \(E_{1x} + E_{2x} = 0 \rightarrow \dfrac{k*q}{r_1^2}\cos{\theta_1} + \dfrac{-k*2q}{r_2^2}\cos{\theta_2}=0\) (1) \(E_{1y} + E_{2y} = 0 \rightarrow \dfrac{k*q}{r_1^2}\sin{\theta_1} + \dfrac{k*2q}{r_2^2}\sin{\theta_2}=0\) (2)
04

Analyzing the equations

Dividing both sides of equation (1) by \(k*q\) and equation (2) by \(k*2q\), we get: \(\dfrac{1}{r_1^2}\cos{\theta_1} = \dfrac{2}{r_2^2}\cos{\theta_2}\) (3) \(\dfrac{1}{r_1^2}\sin{\theta_1} = \dfrac{1}{r_2^2}\sin{\theta_2}\) (4) Now, squaring both equations and adding them, we have: \(\dfrac{1}{r_1^4}\cos^2{\theta_1} + \dfrac{2}{r_1^4r_2^2}\cos{\theta_1}\cos{\theta_2} + \dfrac{4}{r_2^4}\cos^2{\theta_2} = \dfrac{1}{r_1^4}\sin^2{\theta_1} + \dfrac{1}{r_1^2r_2^2}\sin{\theta_1}\sin{\theta_2} + \dfrac{1}{r_2^4}\sin^2{\theta_2}\) Using the identity \(\sin^2{\theta}+\cos^2{\theta}=1\), we get: \(\dfrac{1}{r_1^4} + \dfrac{4}{r_2^4} = \dfrac{3}{r_1^2r_2^2}\) (5)
05

Conclusion

Looking at equation (5), we observe that it is independent of \(\theta_1\) and \(\theta_2\). As a result, we cannot find any solution for this equation that has \(y\neq0\). In other words, there is no point not on the \(x\)-axis where the electric field \(\vec{E}=0\).

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

A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d\) (a) Write an expression for the magnitude of the electric field at a point \((x, 0)\) a large distance \((x \gg d)\) from the midpoint of the charges on a line perpendicular to the dipole axis. [Hint: Use small angle approximations.] (b) Give the direction of the field for \(x>0\) and for \(x<0\).
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
(a) What would the net charges on the Sun and Earth have to be if the electric force instead of the gravitational force were responsible for keeping Earth in its orbit? There are many possible answers, so restrict yourself to the case where the magnitude of the charges is proportional to the masses. (b) If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies would have net charges that are approximately proportional to their masses. Could this possibly be an explanation for the Earth's orbit?
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=d\) (point \(P\) )?
\(\mathrm{A} \mathrm{K}^{+}\) ion and a \(\mathrm{Cl}^{-}\) ion are directly across from each other on opposite sides of a membrane 9.0 nm thick. What is the electric force on the \(\mathrm{K}^{+}\) ion due to the \(\mathrm{Cl}^{-}\) ion? Ignore the presence of other charges.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Fluids

Read Explanation

Electromagnetism

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