Added Notes

August 31, 2005

Definitions Electric Field = Force per unit charge. Types of charge distributions

Point Charges Lines of Charge Areas of Charge Volumes of Charge

General:

unitjeschall

jqk rE arg

2r

Point ChargesIn the Figure, the four particles are fixed in place and have charges q1 = q2

= +5e, q3 = +3e, and q4 = -12e. Distance d = 9.0 mm. What is the

magnitude of the net electric field at point P due to the particles?

                                              

Line of Charge

Need =charge per unit length

unitr

dsk rE

2

ds

dq= ds

r

A Harder Problem

A line of charge=charge/length

setupsetup

dx

L

r

x

dE dEy

2/

02/322

2/

02/322

22

2

2

22

)(2

)(2

)()cos(

)(

)cos(

L

x

L

x

L

Lx

xr

dxkrE

xr

dxrkE

xr

r

xr

dxkE

(standard integral)

Completing the Math

r

kL

r

klE

Lr

L

Lrr

kLE

x

x

2

2

4

:line long VERY a oflimit In the

4

:nintegratio theDoing

22

22

1/r dependence

Surface Charge

Need Surface Charge density = =charge per unit area. unitr

dAk rE

2

dA

r runit

The Geometry

Define surface charge density=charge/unit-area

dq=dA

dA=2rdr

(z2+r2)1/2

dq= x dA = 2rdr

(z2+r2)1/2

R

z

z

rz

rdrzkE

rz

z

rz

drrk

rz

dqkdE

02/322

2/1222222

2

2)cos(

(z2+r2)1/2

Final Result

0z

220

2E

,R

12

When

Rz

zEz

Look at the “Field Lines”

Kinds of continuously distributed charges Line of charge

or sometimes = the charge per unit length. dq=ds (ds= differential of length along the line)

Area = charge per unit area dq=dA dA = dxdy (rectangular coordinates) dA= 2rdr for elemental ring of charge

Volume =charge per unit volume dq=dV dV=dxdydz or 4r2dr or some other expressions we will look at later.

The Sphere

dqr

thk=dr

dq=dV= x surface area x thickness= x 4r2 x dr

Summary

222

,2

2

2

)()()(

r

rdsk

r

rdAk

r

rdVk

r

Qk

q

General

r

Qk

q

r

qQk

unitjj

jjj

unit

unit

E

rF

EE

rF

E

rF

(Note: I left off the unit vectors in the lastequation set, but be aware that they should

be there.)

The figure shows two concentric rings, of radii R and R ' = 3.25R, that lie on the same plane. Point P lies on the central z axis, at distance D = 1.90R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-5.39]   Q

The figure shows a plastic ring of radius R = 46.0 cm. Two small charged beads are on the ring: Bead 1 of charge +2.00 µC is fixed in place at the left side; bead 2 of charge +5.60 µC can be moved along the ring. The two beads produce a net electric field of magnitude E at the center of the ring. At what positive and negative values of angle   should bead 2 be positioned such that E = 2.00   105 N/C? (Measure the angle from the positive x axis taking counterclockwise to be positive.)

In the figure, a thin glass rod forms a semicircle of radius r = 4.00 cm. Charge is uniformly distributed along the rod, with +q = 4.00 pC in the upper half and -q = -4.00 pC in the lower half.

(a) What is the magnitude of the electric field at P, the center of the semicircle?[28.6] N/C(b) What is its direction?[-90]° (counterclockwise from the positive x axis)

d

ds