Announcements (Jan 15)

• “Preliminary" Lecture PDF file will be posted 1 day ahead of class time – Due to a number of requests: I will be posting a super-set of my

lecture PDF for each class a day or two before that class. Naturally there will be more material in it than I will cover because it is difficult to predict where one would finish exactly. Also: I might end up making some changes to the lectures as we go (usually resulting in reordering of some slides and possibly omission of some slides)

• HW#1 now due Wed. Jan 23 5am (because of Monday holiday) – Monday is a holiday so the due time HW#1 has been be

postponed by 24 hours to Wed. Jan 23 at 5am. 1

18.5 Coulomb’s Law

2

Example Vector addition of Electric fields from point charges Assuming the horizontal to be the x-direction, and vertical the y-direction, determine the magnitude and direction of the net electric field at the location of q1.

3

18.5 Coulomb’s Law

18.5 Coulomb’s Law

4

Example Vector addition of Electric fields from point charges Assuming the horizontal to be the x-direction, and vertical the y-direction, determine the magnitude and direction of the net electric field at the location of q1.

( )( )( ) C

N10397.2m15.0

C100.6CmN1099.8 62

6229

212

22 ×=

×⋅×==

−

rq

kE

( )( )( ) C

N 10495.4m10.0

C100.5CmN1099.8 62

6229

213

33 ×=

×⋅×==

−

rq

kE

Solution: First we apply the Inverse Coulomb’s Law for the electric fields to find the electric field magnitudes E2 and E3 due to q2 and q3:

21 j

jjj r

qkEE ==

where r1j represents the distance between q1 and qj

The resulting electric field at the location of q1 is the vector sum of the electric fields created by charges q2 and q3 individually. We make this sum component by component: The two fields point TOWARD the q2 and q3, respectively, as indicated in the figure.

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18.5 Coulomb’s Law

N/C 105.196 N/C 10495.4 N/C 107009.0)0cos()N/C 10495.4()73cos()N/C 10397.2(

coscos

666

o6o63322

×=×+×=

×+×=

++= θθ EEEx

o11-

6262622

24)442.0(tantan

:axis- thefrom from wise-clock-counter of Angle

N/C 107.5)N/C 10293.2()N/C 105.196(

===

+

×=×+×=+=

−

x

y

yx

EE

xE

EEE

θ

E

N/C 10293.2)0sin()N/C 10495.4()73sin()N/C 10397.2(

sinsin

6

o6o6

3322

×=

×+×=

++= θθ EEE y

18.7 Electric Field Lines Electric field lines or lines of force provide a map of the electric field in the space surrounding electric charges.

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Electric field lines are always directed away from positive charges and toward negative charges. You can think of a positive charge as the “source” of field lines

18.7 Electric Field Lines

Electric field lines are always directed away from positive charges and toward negative charges. You can think of a negative charge as the “sink” of field lines

7

The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.

Electric Dipole

18.7 Electric Field Lines

8

If we could, the field lines should be drawn in 3D, where the line density--how closely packed the lines are in terms of (number of lines)/(cross-sectional area)-- is proportional to the field strength (magnitude of the electric field). And the direction of the “flow” gives the direction of the electric field.

18.7 Electric Field Lines

Conceptual Example: Drawing Electric Field Lines There are three things wrong with part (a) of the drawing. What are they?

9

10

Answer: (1) The field lines cross at point P--

they should not cross one another (albeit this is a 2D representation of a 3D situation)

(2) The number of field lines going into the three negative charges are not proportional to their charge magnitude

(3) The field lines between the central 4q charge and the –q charge on the left are parallel and equally spaced the whole way between the two: this would mean that the field strength is uniform in this region. But we should expect the field strength to be weaker in the middle than near the two charges

18.6 The Electric Field

THE PARALLEL PLATE CAPACITOR

Parallel plate capacitor

oo AqE

εσ

ε==

( )2212 mNC1085.8 ⋅×= −οε

surface charge density (charge/area)

11

Electric field lines always begin on a positive charge and end on a negative charge and do not stop in mid-space.

12

vf

s

vi

+Q

-Q

A A

Example: A capacitor has plate area A=4.50m2 and plate separation s=2.00cm. A charge of Q=22.5µC is placed on the left plate, and –Q on the right

A proton (m=1.67x10-27kg) with initial velocity of vi=1.50x106m/s in the horizontal direction when injected into the gap at the left plate. Find its final speed vf as it crashes into the right plate.

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vf

s

vi

+Q -Q

A A

Example: A (vertical) parallel plate capacitor have the following attributes: Area of plates: A=4.50m2 Plate separation: s=2.00cm A charge of Q=22.5µC is placed on the left plate, and –Q on the right A proton (m=1.67x10-27kg) with initial velocity of vi=1.50x106m/s in the horizontal direction when injected into the gap at the left plate. Find its final speed vf as it crashes into the right plate.

2132712

14519

522122

-6

00

m/s 10413.5)kg 1067.1/()N 10040.9(/N 10040.9)N/C 1065.5C)( 1060.1(

N/C 10650.5)mN/(C 1085.8)[m 50.4(

)C10 5.22(

×=××==

×=××==

×=⋅×

×===

−−

−−

−

mFaqEF

AQEεε

σ

To be continued