University Physics: Waves and Electricity Ch22. Finding the Electric Field – I Lecture 7 Dr.-Ing....

University Physics: Waves and Electricity

Ch22. Finding the Electric Field – ILecture 7

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

7/2Erwin Sitompul University Physics: Wave and Electricity

The Electric Field

The Coulomb’s law tells us how a charged particle interacts with another charged particle.

The question now: Since the particles do not touch, how can one particle push or pull the other? How can there be such an action at a distance with no visible connection between the particles?

The concept of Electric Field is introduced to explain this question.


The Electric Field

The electric field is a vector field. It consists of a distribution of vectors, one for each point in the region around a charged object.

We can define the electric field at some point, such at point P, by placing a positive charge q0, called a test charge.

We then measure the electrostatic force F that acts on the test charge.

The electric field E at point P is defined as:

0

FE

q

→

→

• The direction of force defines the direction of field


The Electric Field

The positive test charge q0 does not “see” the charged object. Instead, it “feels” the electric field produced by the charged object and gives response.

The SI unit for the electric field is the newton per coulomb (N/C).

Note: The field at point P existed both before and after the test charge was put there.

Here, we assume that the presence of the test charge does not affect the charge distribution on the charged object.


Electric Field Lines

In order to understand it better, we will try to visualize the electric field now.

Michael Faraday introduced the idea of electric fields in the 19th century and thought of the space around a charged body as filled with electric field lines .

The direction of the field lines indicate the direction of the electric force acting on a positive test charge.

The density of the field lines is proportional to the magnitude of the field.

• The field strength is related to the number of lines that cross a certain unit area perpendicular to the field


The figure shows a positive test charge, placed near a sphere of uniform negative charge.

The electrostatic force points toward the center of the sphere.

The electric field vectors at all points are directed radially toward the sphere.

The spreading of the field lines with distance from the sphere tells us that the magnitude of the electric field (field strength) decreases with distance from the sphere.




Electric field lines extend away from positive charge (where they originate) and toward negative charge (where they terminate).



The figure below shows an infinitely large, nonconducting sheet (or plane) with a uniform distribution of positive charge on one side.

Due to symmetry, some forces will cancel one another. The net electrostatic force on the positive test charge will be

perpendicular to the sheet and point away from it.


The Electric Field Due to a Point Charge

From Coulomb’s law, the electrostatic force due to q1, acting on a positive test charge q0 is:

0 10 102

10

rq q

F kr

00

0

FE

q

1102

10

rq

kr

The electric field due to a point charge q1 is:

The field of a positive point charge is shown on the right, in vector form.

The magnitude of the field depends only on the distance between the point charge (as the field source) and the location where the field is measured.


Electric field is a vector quantity. Thus, the net, or resultant, electric field due to more than one

point charge is the superposition of the field due to each charge.

The net electric field at the position of the test charge, due to n point charges, is:

The Electric Field Due to a Point Charge

0,net 01 02 03 0nE E E E E

31 210 20 302 2 2 2

10 20 30 0

ˆ ˆ ˆ ˆr r r rnn

n

q qq qk k k kr r r r

0,net 021 0

rn

mm

m m

qE k

r


Checkpoint

The figure below shows a proton p and an electron e on an x axis.

What is the direction of the electric field due to the electron at:(a) Point S?(b) Point R?What is the direction of the net electric field at(c) Point R?(d) Point S?

RightwardLeftward

LeftwardRightward

• p and e have the same charge magnitude

• R and S are closer to e than to p.


Example 1

A point source q1 = 20 nC is located at S(1,4). Find the electric field E at P(5,1). All units are in SI.

→

ˆ ˆi 4 jSr

10 102

10

rq

E kr

ˆ ˆ5i jPr

SP P Sr r r ˆ ˆ ˆ ˆ(5i j) (i 4 j) ˆ ˆ4i 3j

2 2(4) ( 3)SPr

5

ˆ SPSP

SP

rr

r

ˆ ˆ4i 3j

5

ˆ ˆ0.8i 0.6 j

• Field at the measurement point

• Source charge

• Vector pointing from source charge to measurement point

2 rSP SP

SP

qE k

r

99

2

(20 10 ) ˆ ˆ8.99 10 (0.8i 0.6 j)(5)

ˆ ˆ5.754i 4.315j N C


Example 2

A point source q1 = 20 nC is located at S(1,4). Determine some points near S, where the magnitude of the electric field E is equal to 30 N/C.

2 rSP SP

SP

qE k

r

2 rSP SP

SP

qE k

r

2S

PSP

qE k

r

E E

r 1

• The magnitude of a vector is the scalar value of the vector itself

• The magnitude of a unit vector always equals 1

99

2

(20 10 )30 8.99 10

SPr

92 9 (20 10 )

8.99 1030SPr

2.448SPr • The points near S must be 2.448 m away from S

• Where are they?


Example 2

x

y

4

1

r = 2.448

• Location of points where the magnitude of E is equal to 30 N/C

q1+

P1

P2

P3

P4

Some points near S with E = 30 N/C are:

P1(3.448,4)P2(1,6.448)P3(–1.448,4)P4(1,1.552)


Checkpoint

The figure here shows four situation in which charged particles are at equal distances from the origin.Rank the situations according to the magnitude of the net electric field at the origin, greatest first.

All the same• Be sure to know

how and why


Example 3

Three particles with charges q1 = 2Q, q2 = –2Q, and q3 = –4Q, each with distance d from the origin, are shown in the figure below. What net electric field E is produced at the origin?

→

1 2 3E E E E

31 210 20 302 2 2

10 20 30

ˆ ˆ ˆr r rqq q

k k kr r r

10ˆ ˆcos( 30 )i sin( 30 ) jr d d

20ˆ ˆcos(150 )i sin(150 ) jr d d

30ˆ ˆcos(210 )i sin(210 ) jr d d


Example 331 2

10 20 302 2 2

10 20 30

ˆ ˆ ˆr r rqq q

E k k kr r r

2 2 2

2 1 1 ( 2 ) 1 1 ( 4 ) 1 1ˆ ˆ ˆ ˆ ˆ ˆ3i j 3i j 3i j2 2 2 2 2 2

Q Q Qk k kd d d

2

8 1 ˆ3 i V m2

Qkd

E

2

6.928i N C

kQ

d


Measuring the Elementary Charge

Millikan oil-drop apparatus

Robert A. Millikan in 1910-1913 devised an apparatus to measure the elementary charge e.

When tiny oil drops are sprayed into chamber A, some become charged, either + or –. They then drift to chamber C.

If switch S is open, battery B has no electrical effect on chamber C. If switch S is closed, the battery causes the plates to be charged and set up a downward-directed electric field E in chamber C.

Negatively charged drop will tend to drift upward, positively charged drop will drift downward.

By timing the motion of oil drops, the effect of charge q can be determined, q = ne, where n is integer and e = 1.60210–19 C.

→


Application: Ink-Jet Printing

Ink-jet printer is an invention to meet the need for high-quality, high-speed printing.

The figure below shows a negatively charged ink drop moving between two conducting plate. A uniform downward-directed electric field E has been set up.

The drop is deflected upward and then strikes the paper at a position that is determined by the magnitudes of E and the charge q of the drop.

→

→


Example 4

An ink drop with a mass m of 1.310–10 kg and a negative charge of magnitude Q = 1.510–13 C enters the region between the plates of an ink-jet printer. The drop initially moves along the x axis with the speed vx = 18 m/s. The length L of each plate is 1.6 cm. The plates are charged and thus produce an electric field at all points between them.Assume that field E is downward directed, is uniform, and has a magnitude of 1.4106 N/C. What is the vertical deflection of the drop at the far edge of the plates? (The gravitational force on the drop is small relative to the electrostatic force acting on the drop and can be neglected.)

→


Example 4

F QE

13 6 ˆ( 1.5 10 )( 1.4 10 j) 7 ˆ2.1 10 j N

Fa

m

7

10

ˆ2.1 10 j

1.3 10

3 2ˆ1.615 10 j m s

• y direction

• y direction

x

xtv

21.6 10

18

48.889 10 s

210 0 2y yy y v t a t

4

3 4 212

(0)(8.889 10 )

(1.615 10 )(8.889 10 )

46.380 10 m

0.638 mm

• Can you locate where the 0.638 mm is?


Three particles are fixed in place and have charges q1 = q2 = +p and q3 = +2p. Distance a = 6 μm.What are the magnitude and direction of the net electric field at point P due to the particles?

Homework 6: Three Particles

191.602 10 Cp 191.602 10 Ce


The figure below shows two charged particles on an x axis, –q = –3.2×10–19 C at x = –3 m and q = 3.2×10–19 C at x = 3 m.What are the (a) magnitude and (b) direction of the net electric field produced at point P at y = 4 m?

Homework 6: Two Particles

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University Physics: Waves and Electricity Ch22. Finding the Electric Field – I Lecture 7 Dr.-Ing....

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