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Question: Two particles of chargesq1and q2 , are separated by distance in Fig. 24-40. The net electric field due to the particles is zero at x = d/4 .With V = 0 at infinity, locate (in terms of ) any point on the x-axis (other than at infinity) at which the electric potential due to the two particles is zero.

Short Answer

Expert verified

Answer:

The net electric potential cannot possibly zero at anywhere (except infinity) on the x-axis.

Step by step solution

01

Given data:

The net electric field due the particles q1 and q2 is zero at d/4 on the x axis. Thus, the fields due to q1 and q2 must be directed opposite to each other at that point otherwise the net electric field will not be zero at that point. This means that the two charges q1 and q2 must have the same sign to the charge either negative or positive.

02

Understanding the concept

The electric field is defined as the Coulombic force per unit positive charge due to another charge.

E=kqr2

Here, k is Coulomb’s constant, q charge of the particle, and r is the distance between the charge and the point where the electric field is located.

The electric potential is expressed as follows.

V=kqr

03

Locate (in terms of  ) any point on the x-axis (other than at infinity) at which the electric potential due to the two particles is zero: 

The net electric field due the particles q1 and q2 is zero at d/4 on the x-axis/ Thus, the field due to and must be directed opposite to each other at that point otherwise the net electric field will not be zero at that point.

This means that two charges q1 and q2 must have the same sign to the charge either negative or positive.

The electric field is a vector quantity. Since the direction of the electric field depends on the sign of the charge. But vector potential is a scalar, so there is no direction for electric potential.

If the two particles have the same charge, then the potentials that they produce will always have the same sign. This means the electric potential at any point on the x axis due to the two charges q1 and q2 never cancel out.

Therefore, the net electric potential cannot possibly zero at anywhere (except infinity) on the x-axis.

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Most popular questions from this chapter

Question: The thin plastic rod shown in Fig. 24-47 has length L = 12.0cmand a non-uniform linear charge density, λ=cxwhere c=28.9pC/m2.With V = 0at infinity, find the electric potential at point P1 on the axis, at distance d = 3.00 cmfrom one end.

(a) Using Eq. 24-32, show that the electric potential at a point on the central axis of a thin ring (of charge q and radius R ) and at distance Z from the ring is,V=14πεoqR2+z2

(b) From this result, derive an expression for the electric field magnitude

E at points on the ring’s axis; compare your result with the calculation of E in Module 22-4.

Figure 24-47 shows a thin plastic rod of length L = 13.5cmand uniform charge 43.6 fC. (a) In terms of distance d, find an expression for the electric potential at point P1. (b) Next, substitute variable xfor dand find an expression for the magnitude of the component Exof the electric field at. (c) What is the direction of Exrelative to the positive direction of the xaxis? (d) What is the value of Exat P1 for x = d = 6.20cm? (e) From the symmetry in Fig. 24-47, determine Eyat P1.

In the situation of Question 7, is the work done by your force positive, negative, or zero if the particle is moved (a) from Ato B, (b) from Ato C, and (c) from Bto D? (d) Rank those moves according to the magnitude of the work done by your force, greatest first.

A solid conducting sphere of radius 3.0cmhas a charge of30nc distributed uniformly over its surface. Let A be a point1.0cm from the center of the sphere,S be a point on the surface of the sphere, and Bbe a point5.0cm from the center of the sphere.What are the electricpotential differences

(a) VSVBand

(b) VAVB?

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