Electric Fields - General Physics II - Lecture Notes | PH 222, Study notes of Physics

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PH 222-3A Spring 2010

ELECTRIC FIELDS

Lectures 2,

Chapter 22 (Halliday/Resnick/Walker, Fundamentals of Physics 8th^ edition)

Chapter 22

Electric Fields

In this chapter we will introduce the concept of an electric field.

As long as charges are stationary, Coulomb’s law describes adequately

the forces among charges. If the charges are not stationary we must use

an alternative approach by introducing the electric field (symbol ).

In connection with the electric field, the following topics will be

covered:

-Calculating the electric field generated by a point charge.

-Using the principle of superposition to determine the electric field

created by a collection of point charges as well as continuous charge

distributions.

-Once the electric field at a point P is known, calculating the electric

force on any charge placed at P.

• Defining the notion of an “electric dipole.” Determining the net force,

the net torque, exerted on an electric dipole by a uniform electric field,

as well as the dipole potential energy.

E

2

Consider the positively charged rod shown in the figure. For every point in the vicinity of the rod we define the electric field vector as follows: We place

P

E

Definition of the Electric Field Vector

0

0

a test charge at point. We measure the electrostatic force exerted on by the charged rod. We define the electric field .

vector at point as: SI Units:

q P F q

E P

E F

q

positiv

e

0

From the definition it follows that is parallel to. We assume that the test charge is small enough so that its presence at point does not affect the charge distribution on the

N/C

rod

E F

q P

Note :

and thus alter the electric field vector E we are trying to determine.

0

E F

q

q

q o r

P

E

2 0

q

E

r

0 0

Consider the positive charge shown in the figure. At point a distance from we place the test charge. The force exerted on by is equal to:

q P r q q q q

Electric Field Generated by a Point Charge

0 0 2 0 0 0 0 2 0 2

The magnitude of is a positive number. In terms of direction, points radially as shown in the figure. If were a negative charge the magnitude of

F q q r E F q q^ q q q r r E E

q

outward

would remain the same. The direction of would point radially instead.

E

E inward

Problem (Electric field). To measure the magnitude of the horizontal electric field, an experimenter attaches a small charged cork ball to a string and suspends this device in the electric field. The electric force pushes the cork ball to one side, and the ball attains equilibrium when the string makes an angle of 35 with the vertical. The mass of the ball is 3x10-5^ kg, and the charge on the ball is 4x10-7^ C. What is the magnitude of the electric field?

5.1x10^2 N/C

X

d

+ q

- q

-q

+ q /2 + q /

A system of two equal charges of opposit e sign placed at a distance apart is known a s an "electric dipole." For every electric dipole we associate a vector known as "the electric

q d

Electric Dipole

dipole moment" (symbol ) defined as follows: The magnitude The direction of is along the line that connects the two charges and points from - to. Many molecules have a built-in elect

p p qd p q q

2

ric dipole moment. An example is the water molecule (H O). The bonding between the O atom and the two H atoms involves the sharing of 10 valence electrons (8 from O and 1 from each H atom).

The 10 valence electrons have the tend ency to remain closer to the O atom. Thus the O side is more negative than th e H side of the H O molecule.^11

Consider the continuous charge distribution shown in the figure. We assume that we know the volume density of the electric charge. This i

Electric Field Generated by a Continuous Charge Distribution

s defined as^3 Our goal is to determine the electric field generated by the distribution at a given point. This type of problem can be solved using t

he

Uni

principle of superpos

dq ts: C/m ). dV dE P ition as described below. Divide the charge distribution into "elements" of volume. Each element has charge. We assume that point is at a distance from. Determine the electric field generated b

dV dq dV P r dq dE

0 2

y at point. The magnitude of is given by the equation. 4 Sum all the contributions: 1 ˆ. 4

dq P dE dE dE dq r E dVr r

P

dq

r

dE

dV r ˆ

13

A C

dq

Determine the electric field generated at point by a uniformly charged ring of radius and total charge. Point lies on the normal to the ring plane that passes through the ring cent

E P

R q P

Example :

2 2

er , at a distance. Consider the charge element of length and charge shown in the figure. The distance between the element and point is. The charge generates at an electric fie

C z dS dq P r z R dq P

0 2

0 3 2 2 3/ 0

ld of magnitude that points outward along the line :

. The -component of is given by 4 cos. From triangle PAC we have: cos / . (^4 )

z z z z

dE AP dE dq z dE r dE dE z r dE zdq^ zdq E dE r (^) z R

2 2 3/2^2 2 3/ (^4 0 )

z E z^ dq zq z R z R 14

2 2 3 2 0 m 2

2

2

1 2 2 1

0

2

2

4 (^ )^ (2 ) To solve this integral, we cast it in the form X by setting ( ), 3 , and (2 ) 2

( ) 4 4 1 4 1

2

2 1

R o

R m m o o o

o

E dE z z r r dr

dX X z r m dX r

z

dr

z z X z z r

E z R

E X dX (^) m

If we let (infinite sheet) while keeping finite, the second term in the parentheses approache

2 (infinite sheet) This is the electic field produced by an infin

s zero and

ite sheet of uniform

o

E

R z

charge located on one side of a nonconductor such as plastic.

P electric field lines

Q

EP Q E

In the 19th century Michael Faraday introduced the concept of electric field lines, which help visualize the electric field vector without using mathematics. For the relatio

E

Electric Field Lines.

n between the electric field lines and E :

1. At any point P the electric field vector E is tangent to the electric field lines.

P

EP electric field line

2. The magnitude of the electric field vector E is proportional to the density of the electric field lines.

E P EQ

q

q

negative charges

Electric field lines extend away from (where they originate) and toward (where they terminate). Example 1 : Electric field lines of a negative p

positive

oint char

charges

ge - q :

2 0

q

E

r

-The electric field lines point toward the point charge. -The direction of the lines gives the direction of. -The density of the lines/unit area increases as the distance from decreases.

E

q In the case of a positive point charge the electric field lines have the same form but they point out

No

w

te :

ard.

The electric field can be represented graphically by field lines