Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Figure 22-26 shows two charged particles fixed in place on an axis. (a) Where on the axis (other than at an infinite distance) is there a point at which their net electric field is zero: between the charges, to their left, or to their right? (b) Is there a point of zero net electric field anywhere offthe axis (other than at an infinite distance)?

Short Answer

Expert verified

a) On the axis, the point is between the charges where their net electric field is zero.

b) There is no off axis points at which the net electric is zero.

Step by step solution

01

Understanding the concept of electric field 

Considering the concept that the electric field lines move toward the negative charge and lines radially move away from the positive charge, the net electric field can be calculated. If the right direction is said to be positive, the net field at point P can be calculated.

Formula:

The electric field due to a charge at a point, E=kqr2 (i)

02

a) Calculation of the location of the point where net field is zero

For the condition of having a point at which the net field is zero, the distance at which the net field is zero can be calculated using equation (i) as follows:

E+q+E3q=0k(+q)d12+k(3q)d22=01d12=3d22d2=3d1

Now, consider that the point is at the right hand side of both the charges, and then the point will be at a distanced2from the charge+qwhich will not be three times ofd1from charge 3q. Thus, there cannot be any point right to both the charges.

Now, considering the point at the left of both the charges will contradict the above calculation case as field line from positive charge will be radially outward from the charge towards the point that implies the net field will subtract each other.

Hence, the point is between the charges on the axis.

03

b) Calculation to check the existence of off-axis points 

As the charges are of different magnitudes, their net electric fields at any other axis will be due to the contribution of their x and y-components respectively.

Hence, there will be no off-axis points where the net field value will be zero.

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 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

In Fig. 22-34 the electric field lines on the left have twice the separation of those on the right. (a) If the magnitude of the field at Ais40N/C, what is the magnitude of the force on a proton at A? (b) What is the magnitude of the field at B?

Question: A 10.0 g block with a charge of+8.00×10-5Cis placed in an electric field E=3000i^-600j^N/C. What are the (a) magnitude and (b) direction (relative to the positive direction of thexaxis) of the electrostatic force on the block? If the block is released from rest at the origin at time t = 0, what is its (c) x and (d)ycoordinates at t =3.00s?

A certain electric dipole is placed in a uniform electric field of magnitude40N/C

. Figure 22-63 gives the magnitude t of the torque on the dipole versus the anglebetween field and the dipole moment .The vertical axis scale is set byts=1.00×1028N.m. What is the magnitude ofp?

Find an expression for the oscillation frequency of an electric dipole of dipole momentpand rotational inertia Ifor small amplitudes of oscillation about its equilibrium position in a uniform electric field of magnitude E.

Two charged beads are on the plastic ring in Fig. 22-44a. Bead 2, which is not shown, is fixed in place on the ring, which has radiusR=60.0 cm. Bead 1, which is not fixed in place, is initially on the x-axis at angleθ=0°. It is then moved to the opposite side, at angleθ=180°, through the first and second quadrants of the x-ycoordinate system. Figure 22-44bgives the xcomponent of the net electric field produced at the origin by the two beads as a function of, and Fig. 22-44cgives the ycomponent of that net electric field. The vertical axis scales are set by Exs=5.0×104 N/CandEys=9.0×104 N/C. (a) At what angle u is bead 2 located? What are the charges of (b) bead 1 and (c) bead 2?

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.

Sign-up for free