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Chapter 3: Q3.7P (page 129)

Find the force on the charge +qin Fig. 3.14. (The xyplane is a grounded conductor.)

Short Answer

Expert verified

Answer

The net force on q is -14蟺蔚0(29272d2)z.

Step by step solution

01

Given data

Consider the given figure as shown below.

Here, from the given figure it is clear that xyplane is grounded conductor, thus potential is zero

02

Determine force

As +qinduces an equal and opposite charge of -qat a distance of Z=-3dand -2qinduces +2qat distance of Z=-d.

Write the expression for force on +qdue to -2q,.

F1=14蟺蔚0q(-2q)(2d)2z 鈥︹�� (1)

Write the expression for force on +qdue to +2q,.

F1=14蟺蔚0q(-2q)(4d)2z 鈥︹�� (2)

Write the expression for force on +qdue to -q.

F1=14蟺蔚0q(-2q)(6d)2z 鈥︹�� (3)

03

Determine the net force

Write the expression for the net force on q .

F=F1+F2+F3=q4蟺蔚0[-2q2d2+2q4d2+-q6d2]z=q24蟺蔚0[-12+18-136]z=q24蟺蔚0(29q272d2)z

Thus, the net force on q is -q24蟺蔚0(29q272d2)z.

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Most popular questions from this chapter

a) Using the law of cosines, show that Eq. 3.17 can be written as follows:

Vr,胃=14蟺蔚0qr2+a2-2谤补肠辞蝉胃-qR2+raR2-2谤补肠辞蝉胃

Where rand 胃are the usual spherical polar coordinates, with the z axis along the

line through q. In this form, it is obvious that V=0on the sphere, r=R.

b) Find the induced surface charge on the sphere, as a function of 胃. Integrate this to get the total induced charge. (What should it be?)

c) Calculate the energy of this configuration.

A stationary electric dipole p鈬赌=pz^is situated at the origin. A positive

point charge q(mass m) executes circular motion (radius s) at constant speed

in the field of the dipole. Characterize the plane of the orbit. Find the speed, angular momentum and total energy of the charge.

Show that the average field inside a sphere of radius R, due to all the charge within the sphere, is

Eave=-14蟺蔚0蚁R3

Where 蚁is the total dipole moment. There are several ways to prove this delightfully simple result. Here's one method:

(a) Show that the average field due to a single chargeqat point r inside thesphere is the same as the field at r due to a uniformly charged sphere with

蚁=q/(43蟺R3), namely

14蟺蔚0(43蟺R3)鈭�qr2rd味'

Where r is the vector from r to d味

(b) The latter can be found from Gauss's law (see Prob. 2.12). Express the answerin terms of the dipole moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

(d) While you're at it, show that the average field over the volume of a sphere, dueto all the charges outside, is the same as the field they produce at the center.

In Prob. 2.25, you found the potential on the axis of a uniformly charged disk:

V(r,0)=蟽2蔚0(r2+R2鈭�r)

(a) Use this, together with the fact that Pi(1)=1to evaluate the first three terms

in the expansion (Eq. 3.72) for the potential of the disk at points off the axis, assuming r>R.

(b) Find the potential for r<Rby the same method, using Eq. 3.66. [Note: You

must break the interior region up into two hemispheres, above and below the

disk. Do not assume the coefficients A1are the same in both hemispheres.]

Two long straight wires, carrying opposite uniform line charges,卤Aare situated on either side of a long conducting cylinder (Fig. 3.39). The cylinder(Which carries no net charge) has radius ,and the wires are a distance from the axis. Find the potential.
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