Homework 3

Ch. 24: 19, 25, 29, 43, 45Ch. 25: 8, 10

19)

Gauss' s Law : 0 = qenc

a) The electric field at points outside the conducting shell

The total charge enclosed by a Gaussiancylinder enclosing the conducting shell and the rod is :

qenc = qrod + qshell = H+qL + H2 qL = qThe electric field will point outward radially from theshell and be perpendicular to the surface in every direction. Thus,the angle between the field and the normal vector to the surfaceis zero so the flux through the Gaussian cylinder will be :

= E A cos 0 = E AThe flux through the two ends of the cylinder will be zero because the electric fieldis parallel to the surface leaving only the cylindrical surface area A = 2 r h

We choose the cylinder to be of height h =L so that it just encloses the shell and rod lengthwise.Putting this all into Gauss' s Law, we get :

0 E 2 r L = q

Solving for E, we get :

E =q

2 0 r L

for any place outside the shell in the radial direction at distance r

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b) The distribution of charge on the shell

The positively charged rod within the shell will attractcharges onto the inside surface of the shell with net charge q

This leaves a net charge of q on the outside surface of the shellTotal charge in the shell is conserved :

qtotal = qinner + qouter = HqL + HqL = 2 q c) The electric field in the region between the shell and the rod

Between the shell and the rod, the total enclosed charge is simply the rod.Thus, qenc = +q

The same type of Gaussian cylinder in part aL is used here soE =

+q2 0 r L

for any place between the shell and rod

25)

The infinitely long cylinder of radius R will be similar to theinfinitely long wire except that instead of a linear charge density ,

we will have a volume charge density JReminder : = chargevolume

N a) inside the cylinder (r

same direction everywhere on that part of the Gaussian cylinder.Applying Gauss' s Law Hqenc = 0 L, we get :

r2 h = 0 E 2 r h

Solving for E, we get

E = r2 0

b) outside the charged cylinder (r>R)

The same rules to drawing our Gaussian cylinder apply for outsidethe charged cylinder except now the enclosed charge is not the radius rof the Gaussian cylinder but the radius R of the charged cylinder sincethe charge is only located inside the infinitely long charged cylinderqenc = R2 hThe flux is the same H = E A = E 2 r hL as is Gauss' s law Hqenc = 0 L, so we get : R2 h = 0 E 2 r hThis time solving for E, R and r do not cancel , so

E = R2

2 0 r

29)

For this ball to be holding in a position where the thread makes an

angle of 30 with the sheet, the tension created in the thread from theelectrical repulsion between the ball and the sheet must balance thegravitational pull. Let T be the magnitude of the tension in the thread.

Using the component version of the equilibrium equation, Fx = 0 Fy = 0

The forces in the x direction are the Coulomb force FC and the x component of the tension in the thread

Fx = FC + Tx = 0

The forces in the y direction are the gravitational force Fg and the y component of the tension

Fy = Fg + Ty = 0

The x and y components of the tension are related to the magnitude of the tension andthe angle it makes with the axis by the usual vector decomposition equations :

Tx = T sin Ty = T cos

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The gravitational force near the earth is given by the usualFg = m gHnegative since it points in the negative y directionLThe Coulomb force is related to the electric field by :

Fc = q Eand the electric field due to an infinitecharged sheet is derived using Gauss' s law and is given by :

E =

2 0Thus,

Fc =q 2 0

Combining these expressions we getHwe add a negative sign in front of Tx since it points in the negative x directionL :q 2 0

T sin = 0

m g + T cos = 0Solving each of these for the common term T, we get :

T = q 2 0

1sin

T = mgcos

Equating the two T' s and solving for Hwhat the problem asksL, we get : =

2 0 m g tan q

Now plugging in the values for the variables,

0 = 8.851012 Fm

ikjjjj Fm = CV 1m = CJC 1m = C2N m 1m = C2kg ms2 m 1m = C2 s2kg m3 y{zzzzq = 2.0108 Cm = 1.0 mg H = 1.0103 g = 1.0106 kgL

g = 9.81 ms2

= 30

We get

= 5.0109C2 s2kg m3 kg

ms2

C= 5.0

nCm2

(2 significant figures since the values of m and g are given to two significant figures)

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43)

a) Electric field within the sphere (r

c) Electric field inside the shell (b

45)

=Ar'

for the spherical shell where r' is the amount of r within the shell

To find the amount of charge in theshell enclosed by a Gaussian sphere with radius Ha < r < bL,we need to add up all of the infintessimal shells of volume multipliedby their charge density Hthis is necessary because is not constantand depends on how much of the volume has been added in alreadyL

qshellenc = dVsince the volum e of the shell is V =

43

r'3

dVdr'

= 4 r'2 or dV = 4 r'2 dr'

Putting this and the definition of into the integral,setting the limits of integration from the inner surface of the shell at r =a to the radius of the Gaussian sphere r

qshellenc = a

r Ar'

4 r'2 dr' = 4 A a

r

r' dr' = 4 A12

Hr2 a2LSo, the enclosed charge by a Gaussian sphere of r > a is :

qenc = qpoint + qshellenc = H+qL + 2 A Hr2 a2LBy Gauss' s Law,0 E 4 r2 = q + 2 A Hr2 a2LSolving for E,

E = 14 0

ikjj qr2 + 2 A 2 A a2r2 y{zzIn order for E to be uniform throughout the shell,

it cannot depend upon r so the terms that depend on r must cancelqr2

2 Aa2

r2= 0

so A must equal :

A =q

2 a2

8)

E HrL = qr4 0 R3

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a)

by definition,

V = i

f

E. d s

in spherical coordinates, d s

= dr r + r d

+ r sin d

since Epoints only in the direction of r

E. d s

=

q r4 0 R3

dr

choosing the initial point as r =0 and the final point as some arbitrary radius r', we get :

V = 0

r' q r4 0 R3

dr =q

4 0 R3 0r'r dr = q4 0 R3 12 Hr'2 02L=

q8 0 R3

r'2

b)

The difference in the electric fieldbetween a point on the surface and the sphere' s center is : V = V Hr' = RL V Hr' = 0L =

q8 0 R3

R2 q

8 0 R3 02 = q

8 0 R3R2 = q

8 0 R

c)

if q >0, the potential is greater inside the sphere than outV(r'= R) < V(r'

10)

a)

Eq. 25 18 : Vf Vi = i

f

E. d s

Eq. 24 13 : E =

2 0For an infinite sheet, E points perpendicular to the surface,which in our case is the z direction.In three dimensional, rectangular Cartesian coordinates,

d s

= dx x + dy y + dz z

Thus, E. d s

=

2 0

dz

So, V HzL V Hz = 0L = 0

z

2 0 dz =

2 0 0zdz = 2 0 Hz 0L

V Hz = 0L = V0Thus,

V HzL = V0 2 0 z b)

The amount of work done by moving an object of charge q through a potential is :

Wq

= Vf Vi

In our case,Wq

= JV0 2 0 z N V0So,

W = q2 0

zikjjjjNote : the units are correct for work Cm2 CC2N m2 m = N m = Jy{zzzz

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