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In Fig. 22-27, two identical circular non-conducting rings are centered on the same line with their planes perpendicular to the line. Each ring has charge that is uniformly distributed along its circumference. The rings each produce electric fields at points along the line. For three situations, the charges on rings Aand Bare, respectively, (1)q0andq0, (2)-q0and-q0, and (3)-q0and.q0Rank the situations according to the magnitude of the net electric field at (a) pointP1midway between the rings, (b) pointP2at the center of ring B, and (c) pointP3to the right of ring B, greatest first.

Short Answer

Expert verified

a)The rank of the situations according to the magnitude of net field at pointP1 is|E(3)|>|E(1)|=|E(2)| .

b) The rank of the situations according to the magnitude of net field at point P2is |E(1)|=|E(2)|=|E(3)|.

c) The rank of the situations according to the magnitude of net field at pointP3 is|E(1)|=|E(2)|>|E(3)|.

Step by step solution

01

Understanding the concept of electric field of a ring

The electric field is directed towards the negative charge and away from the positive charge. If the right direction is said to be positive, the net field can be calculated at all the given axial points using the field formula.

The electric field of a ring at an axial point,E=kqx(x2+a2)3/2 (i)

where,xis the distance of the point from the center of the ring that lies on the axis andis the radius of the ring.

02

a) Calculation of the rank according to magnitude of electric fields at point P1

Let, be the distance between the two rings bed.

As the point is located in between the two ringsB, then the net electric filed due to the charges q0andq0at the midway between the rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

E(1)=8kq(d/2)(d2+4a2)3/28kq(d/2)(d2+4a2)3/2(Asthepointisatdistanced/2fromeachring.)|E(1)|=0

Now, the net electric filed due to the charges atq0andq0the midway between the rings can be given using equation (i) as follows:

E(2)=8kq(d/2)(d2+4a2)3/2+8kq(d/2)(d2+4a2)3/2(Asthepointisatdistanced/2fromeachring.)|E(2)|=0

Now, the net electric filed due to the charges q0andq0at the midway between the rings can be given using equation (i) as follows:

E(3)=8kq(d/2)(d2+4a2)3/28kq(d/2)(d2+4a2)3/2(Asthepointisatdistanced/2fromeachring.)=8kqd(d2+4a2)3/2|E(3)|=8kqd(d2+4a2)3/2

Hence, the required rank is.|E(3)|>|E(1)|=|E(2)|

03

b) Calculation of the rank according to magnitude of electric fields at point P2

Let, be the distance between the two rings be d.

As the point is located at the center of ring B, then the net electric filed due to the chargesq0andq0at the center of ring B can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

E(1)=kqd(d2+a2)3/2kq(0)(02+a2)3/2|E(1)|=kqd(d2+a2)3/2

Now, the net electric filed due to the chargesq0andq0 at thecenter of ring B can be given using equation (i) as follows: (Since, both the charges are positive, the field is radially towards them)

E(2)=kqd(d2+a2)3/2kq(0)(02+a2)3/2|E(2)|=kqd(d2+a2)3/2

Now, the net electric filed due to the chargesq0andq0at the center of ring B can be given using equation (i) as follows:

E(3)=kqd(d2+a2)3/2+kq(0)(02+a2)3/2|E(3)|=kqd(d2+a2)3/2

Hence, the required rank is|E(1)|=|E(2)|=|E(3)| .

04

c) Calculation of the rank according to magnitude of electric fields at point P3 

Let, be the distance between the two rings beand the point be at a distancexfrom ring 2. Thus, the point distance from ring A will be(d+x).

As the point is located right of both rings, then the net electric filed due to the chargesq0andq0at the right of both rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

|E(1)|=kq(d+x)((d+x)2+a2)3/2+kqx(x2+a2)3/2

Now, the net electric filed due to the chargesq0andq0at the right of both rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially towards them)

E(2)=kq(d+x)((d+x)2+a2)3/2kqx(x2+a2)3/2|E(2)|=kq(d+x)((d+x)2+a2)3/2+kqx(x2+a2)3/2

Now, the net electric filed due to the chargesq0andq0at the right of both rings can be given using equation (i) as follows:

E(3)=kq(d+x)((d+x)2+a2)3/2+kq(x)(x2+a2)3/2|E(3)|=kq(d+x)((d+x)2+a2)3/2kq(x)(x2+a2)3/2

Hence, the required rank is |E(1)|=|E(2)|>|E(3)|.

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Most popular questions from this chapter

Figure 22-38ashows two charged particles fixed in place on an x-axis with separation L. The ratio q1/q2 of their charge magnitudes is . Figure 22-38bshows the xcomponentEnet,Xof their net electric field along the x-axis just to the right of particle 2. The x-axis scale is set by xs=30.0cm. (a) At what value ofx>0isEnet,x is maximum? (b) If particle 2 has charge -q2=-3e, what is the value of that maximum?

An electric dipole with dipole momentp=(3.00i^+4.00j^)(1.24×1030C.m)is in an electric fieldE=(4000N/C)i^(a) What is the potentialenergy of the electric dipole? (b) What is the torque acting on it?(c) If an external agent turns the dipole until its electric dipole moment isp=(4.00i^+3.00j^)(1.24×1030C.m)how much work is done by the agent?

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?

Figure 22-45 shows an electric dipole. What are the (a) magnitude and (b) direction (relative to the positive direction of the xaxis) of the dipole’s electric field at point P, located at distance,r>>d?

Figure 22-22 shows three arrangements of electric field lines. In each arrangement, a proton is released from rest at point Aand is then accelerated through point Bby the electric field. Points Aand Bhave equal separations in the three arrangements. Rank the arrangements according to the linear momentum of the proton at point B, greatest first.

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