In Fig. 22-65, eight particles form a square in which distanced=2.0cm. The charges are,q1=+3e, q2=+e, q3=5e,q4=2e, q5=+3e, q6=+e,q7=5eand q8=+e. In unit-vector notation, what is the net electric field at the square’s center?

Short Answer

Expert verified

The net electric field at the square’s centre is.(1.08×105N/C)i^

Step by step solution

01

The given data

  1. Eight charged particles form a square in which distance, d=2.0cm(as shown in fig.).
  2. The values of the charges,q1=+3e,,q2=+e,q1=+3e,q3=5e,q4=2e,q5=+3eq6=+e,q7=7eandq8=+e.
02

Understanding the concept of the electric field

Using the basic concept of the electric field on a point due to a particle at other points, we can get the individual electric fields of the charges. Then, adding them up will give us the net electric field of the charges on the point.

Formula:

The electric field at a point to a charge,E=q4πεor2r^ (i)

Where, r = The distance of field point from the charge

q = charge of the particle

03

Calculation of the net electric field

Most of the individual fields, caused by diametrically opposite charges, will cancel, except for the pair that lie on the x axis passing through the center. This pair of charges produces a field pointing to the right which is give n using equation (i) as:

E=3e4πεod2i^=3(8.99×109N.m2/C2)(1.6×1019C)(0.020m)2i^=(1.08×105N/C)i^

Hence, the value of the electric field is.(1.08×105N/C)i^

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Most popular questions from this chapter

Two large parallel copper plates are5.0cmapart and have a uniform electric field between them as depicted in Fig. 22-60. An electronis released from the negative plate at the same time that a proton is released from the positive plate. Neglect the force of the particles on each other and find their distance from the positive plate when they pass each other. (Does it surprise you that you need not know the electric field to solve this problem?)

In Fig. 22-56, a “semi-infinite” non-conducting rod (that is, infinite in one direction only) has uniform linear charge density l. Show that the electric field Epat point Pmakes an angle of45°with the rod and that this result is independent of the distance R. (Hint:Separately find the component ofEpparallel to the rod and the component perpendicular to the rod.)

A particle of chargeq1is at the origin of an xaxis. (a) At what location on the axis should a particle of charge4q1be placed so that the net electric field is zero atx=2.0mmon the axis? (b) If, instead, a particle of charge+4q1is placed at that location, what is the direction (relative to the positive direction of the xaxis) of the net electric field atx=2.0mm?

Figure 22-40 shows a proton (p) on the central axis through a disk with a uniform charge density due to excess electrons. The disk is seen from an edge-on view. Three of those electrons are shown: electron ecat the disk center and electrons esat opposite sides of the disk, at radius Rfrom the center. The proton is initially at distance z=R=2.00 cmfrom the disk. At that location, what are the magnitudes of (a) the electric field Ec due to electron ecand (b) the netelectric field Es,net due to electrons es? The proton is then moved to z=R/10.0. What then are the magnitudes of (c) Ecand Es,net (d) at the proton’s location? (e) From (a) and (c) we see that as the proton gets nearer to the disk, the magnitude of Ecincreases, as expected. Why does the magnitude of Es,net from the two side electrons decrease, as we see from (b) and (d)?

In Fig. 22-51, two curved plastic rods, one of charge +qand the other of charge-q, form a circle of radius R=8.50 cm in an x-yplane. The xaxis passes through both of the connecting points, and the charge is distributed uniformly on both rods. If q=15.0 pC, what are the (a) magnitude and (b) direction (relative to the positive direction of the xaxis) of the electric field Eproduced at P, the center of the circle?

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