Four particles (one of charge q,one of charge 3q,and two of charge -2q)are placed as shown in Fig. 3.31, each a distance from the origin. Find a

simple approximate formula for the potential, valid at points far from the origin.

(Express your answer in spherical coordinates.)

Short Answer

Expert verified

The potential at the origin for the point charge is V=12πε02qacosθr2.

Step by step solution

01

Define function

The dipole moment is the product of charge and distance of separation charge from the origin.

Write the expression for dipole moment.

p=∑i=1nqiri ……. (1)

Here, is the charge of ithparticle,ri is the distance of separation charge form the origin.

02

Given data

The position of particle is shown in figure.

It consist of charges as q ,3q-2q,-2q. The separation between the charges denoted by a.

03

Determine potential

Write the expression for the total dipole moment due to four charges.

P=3qaz^-qaz^+-2qay^+-2qa-y^=2qaz^+2qay^+2qay^.........(2)=2qaz^

Write the expression for the position of the point in spherical coordinates.

x=rsinθcosθq=rsinθsinθz=rcosθ

The scalar product of dipole P and position vector r^is,

P.r^ ......(3)

Substitute 2qaz^ for P in equation (3).

P.r^=2qaz^.r^=2qaz^.r^=2qar^cosθ.r^

Substitute r^.r^=1

Thus,

P.r^=2qacosθ.......(4)

Write the equation for potential at distance rdue to point charge q.

V=14πε0P.r^r2......(5)

Substitute 2qacosθ for P .r^, in equation (5)

V=14πε02qacosθr2

Thus, the potential at the origin for the point charge is V=14πε2qacosθr2.

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Most popular questions from this chapter

Buckminsterfullerine is a molecule of 60 carbon atoms arranged

like the stitching on a soccer-ball. It may be approximated as a conducting spherical shell of radius R=3.5A°. A nearby electron would be attracted, according to Prob. 3.9, so it is not surprising that the ion C60-exists. (Imagine that the electron on average-smears itself out uniformly over the surface.) But how about a second electron? At large distances it would be repelled by the ion, obviously, but at a certain distance r (from the center), the net force is zero, and closer than this it would be attracted. So an electron with enough energy to get in that close should bind.

(a) Find r, in A°. [You'll have to do it numerically.]

(b) How much energy (in electron volts) would it take to push an electron in (from

infinity) to the point r? [Incidentally, the C60-ion has been observed.]

Find the potential outside an infinitely long metal pipe, of radius R, placed at right angles to an otherwise uniform electric field E0. Find the surface charge induced on the pipe. [Use your result from Prob. 3.24.]

A spherical shell of radius carries a uniform surface charge on the "northern" hemisphere and a uniform surface charge on the "southern "hemisphere. Find the potential inside and outside the sphere, calculating the coefficients explicitly up to and .

(a) Suppose a charge distribution p1(r⇀)produces a potential V1(r⇀), and some othercharge distribution p2(r⇀)produces a potential V2(r⇀). [The two situations mayhave nothing in common, for all I care-perhaps number 1 is a uniformlycharged sphere and number 2 is a parallel-plate capacitor. Please understand that p1and p2 are not present at the same time;we are talking about two differentproblems,one in which only p1is present, and another in which only p2 ispresent.] Prove Green's reciprocity theorem:

∫allspacep1V2dτ=∫allspacep2V1dτ

[Hint:Evaluate ∫E→1XE→2dτtwo ways, first writing E→1=-∇→V1and using integrationby parts to transfer the derivative to E→2, then writing E→2=-∇→V2and transferring the derivative to E→1.]

(b) Suppose now that you have two separated conductors (Fig. 3.41). If you chargeup conductor by amount Q(leaving uncharged), the resulting potential of bis, say,Vab.On the other hand, if you put that same charge on conductor (leaving uncharged), the potential of would be.Use Green's reciprocitytheorem to show that Vab=Vba(an astonishing result, since we assumed nothingabout the shapes or placement of the conductors).

In Ex. 3.2 we assumed that the conducting sphere was grounded ( V=0). But with the addition of a second image charge, the same basic modelwill handle the case of a sphere at any potentialV0 (relative, of course, to infinity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.

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