ELECTROSTATICS: The study of the behavior of stationary charges ELECTRIC CHARGE

ELECTROSTATICS:The study of the behavior of stationary charges

ELECTRIC CHARGE

There are two types of electric charge, arbitrarily called positive and negative. Rubbing certain electrically neutral objects together (e.g., a glass rod and a silk cloth) tends to cause the electric charges to separate. In the case of the glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains negative charge and therefore becomes negatively charged. After separation, the negative charges and positive charges are found to attract one another.

When a rubber rod is rubbed against fur, electrons When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod.are removed from the fur and deposited on the rod.

The rod is said to beThe rod is said to be negatively chargednegatively charged because of because of anan excessexcess of electrons. The fur is said to be of electrons. The fur is said to be positively chargedpositively charged because of abecause of a deficiencydeficiency of of electrons.electrons.

Electrons Electrons move from move from fur to the fur to the rubber rod.rubber rod.

positive

negative

+ + + +

--

--

When a glass rod is rubbed against silk, electrons are When a glass rod is rubbed against silk, electrons are removed from the glass and deposited on the silk.removed from the glass and deposited on the silk.

The glass is said to beThe glass is said to be positivelypositively chargedcharged because because of aof a deficiencydeficiency of electrons. The silk is said to be of electrons. The silk is said to be negatively chargednegatively charged because of anbecause of an excessexcess of of electrons.electrons.

Electrons Electrons move from move from glass to the glass to the silk cloth.silk cloth.

positive

negative- - - -

+ +

+ +

silk

glass

Pith-ball Electroscope

Gold-leaf Electroscope

Laboratory devices used to study the existence of two kinds of electric charge.

Laboratory devices used to study the existence of two kinds of electric charge.

1. Charge the rubber rod by rubbing against fur.1. Charge the rubber rod by rubbing against fur.

2. Transfer electrons from rod to each pith ball.2. Transfer electrons from rod to each pith ball.

The two negative charges repel each other.

The two negative charges repel each other.

1. Charge the glass rod by rubbing against silk.1. Charge the glass rod by rubbing against silk.

2. Touch balls with rod. Free electrons on the balls 2. Touch balls with rod. Free electrons on the balls move to fill vacancies on the cloth, leaving each of move to fill vacancies on the cloth, leaving each of the balls with a deficiency. (Positively charged.)the balls with a deficiency. (Positively charged.)

The two positive charges The two positive charges repelrepel each each other.other.

The two positive charges The two positive charges repelrepel each each other.other.

fur

Rubber

Attraction

Note that the negatively chargedNote that the negatively charged (green)(green) ball is ball is attractedattracted to the positively chargedto the positively charged (red)(red) ball.ball.

Opposite Charges Attract!Opposite Charges Attract!

silk

glass

Like charges repel; unlike charges attract.Like charges repel; unlike charges attract.

NegNeg PosNegPosPos

Charging by Contact  

Some electrons leave rod and spread over sphere.

Charging by Induction    

Rod does not touch sphere.  It pushes electrons out of the back side of the sphere and down the wire to ground.  The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed.

The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero." Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be transferred from one object to another.

Charged comb attracts neutral bits of paper.

Charged comb attracts neutralwater molecules.

Applications of Electrostatic Charging                                        

Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint.

Negatively charged paint adheres to positively charged metal.

TheThe quantity of chargequantity of charge ((qq)) can be defined in terms can be defined in terms of the number of electrons, but theof the number of electrons, but the Coulomb (C)Coulomb (C) is is a better unit for later work. a better unit for later work.

The Coulomb: 1 C = 6.25 x 1018 electronsThe Coulomb: 1 C = 6.25 x 1018 electrons

The charge on a single electron is:The charge on a single electron is:

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

The The coulombcoulomb (selected for use with electric (selected for use with electric currents) is actually a currents) is actually a very large unitvery large unit for static for static electricity. It is common to use the metric prefixes.electricity. It is common to use the metric prefixes.

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

COULOMB’S LAW Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) between their centers. The equation is:

F = electrostatic force (N)q = charge (C)k = 9x109 N•m2/C2

r = separation between charges (m)

1 22

q qF k

r

The value of k can also be expressed in terms of the permittivity of free space (εo):

ko

1

49x109 N. m2/C2

The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by dividing k by the dielectric constant (K). For a vacuum K = 1, for distilled water K = 80, and for wax paper K = 2.25

Problem-Solving Strategies

1. 1. Draw Draw and and labellabel a figure indicating a figure indicating positive and negative charges along positive and negative charges along with the given distances.with the given distances.

2. Draw the force of 2. Draw the force of attractionattraction or or repulsionrepulsion on the given charge on a on the given charge on a neat, labeled FBD.neat, labeled FBD.

3. Find the 3. Find the resultant force.resultant force.

Important: Important: Do not use the signsDo not use the signs of the of the charges when applying Coulomb's charges when applying Coulomb's law!law!

11.1 Two charges q1 = - 8 μC and q2= +12 μC are placed 120 mm

apart in the air. What is the resultant force on a third charge q3 = - 4 μC placed midway between the other charges?

q1 = - 8x10-6 C

q2= +12x10-6 C

q3 = - 4x10-6 C

r = 0. 120 m

-q1

+q2

-q3

F1

F2

FR

6 69

2

(8 10 )(4 10 )9 10

(0.06)

x xx

1 31 2

q qF k

r

2 32 2

q qF k

r

6 69

2

(12 10 )(4 10 )9 10

(0.06)

x xx

0.06 m 0.06 m

= 80 N

= 120 N

FR = 80 + 120 = 200 N, to the right

11.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are

arranged as shown. Find the resultant force on q3 due to the other

two charges. q1 = +4x10-9 C

q2= -6x10-9 C

q3 = -8x10-9 C

F2

F1

37˚ θ

= 2.88x10-5 N

= 6.75x10-5 N

FR

9 991 3

1 2 2

(4 10 )(8 10 )9 10

(0.1)

q q x xF k x

r

9 992 3

2 2 2 2

(6 10 )(8 10 )9 10

(8 10 )

q q x xF k x

r x

F2

F1

37˚ θFrom the FBD:

Σ Fx = F2 - F1 cos 37˚

= (6.75x10-5) - (2.88x10-5)(cos 37˚)= 4.45x10-5 N

Σ Fy = F1 sin 37˚

= (2.88x10-5)(sin 37˚)= 1.73x10-5 N

F x x ( . ) ( . )4 45 10 173 105 2 5 2 = 4.8x10-5 N

tan.

.1

5

5

173 10

4 45 10

x

x

θ = 21˚

FR (4.8x10-5 N, 21˚)

FR

ELECTRIC FIELD

An electric field is said to exit in a region of space in which an electric charge will experience an electric force. The magnitude of the electric field intensity is given by:

EF

q Units: N/C

The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge would move if it were placed at that point. The electric field lines or lines of force indicate the direction. The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.

11.3 The electric field intensity between two plates is constant and directed downward. The magnitude of the electric field intensity is 6x104 N/C. What are the magnitude and direction of the electric force exerted on an electron projected horizontally between the two plates?

E = 6x104 N/Cqe = 1.6x10-19 C

F = qE = 1.6x10-19 (6x104) = 9.6x10-15 N, upward

11.4 Show that the gravitational force on the electron of example 16-3 may be neglected.

me = 9.11x10-31 kg

FG = mg

= 9.11x10-31 (9.8) = 8.92x10-30 N

The electric force is larger than the gravitational force by a factor of 1.08x1015!

The electric field intensity E at a distance r from a single charge q can be found as follows:

Units: N/C2

k qE

r

11.5 What is the electric field intensity at a distance of 2 m from a charge of -12 μC?

r = 2 mq = -12 μC

Ekq

r

2

9 10 12 10

2

9 6

2

x x( )= 27x103 N/C, towards q

When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from each charge.

Units: N/C

2

k qE

r

11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,

as shown in the figure. Determine the electric field a. At point A

q1 = -6 x10-9 C

q2 = +6 x10-9 C

E1

E2

ER

Ekq

r

2

Ex x

x1

9 9

2 2

9 10 6 10

4 10

( )

( )= 3.38x104 N/C, left

Ex x

x2

9 9

2 2

9 10 6 10

8 10

( )

( )= 8.44x103 N/C, left

E1

E2

ER

ER = E1 + E2

= 3.38x104 + 8.44x103

= 4.22x104 N/C, left

11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,

as shown in the figure. Determine the electric field b. At point B

q1 = -6x10-9 C

q2 = +6x10-9 C

E1

37º

E2

ER

θ

Ex x

x1

9 9

2 2

9 10 6 10

9 10

( )

( )= 6.67x103 N/C

Ex x

x2

9 9

2 2

9 10 6 10

15 10

( )

( )= 2.4x103 N/C

E1

37º

E2

ER

θ

From vector diagram:

Σ Ex = - E2cos 37˚

= - (2.4x103)(cos 37˚)= -1916.7 N/C

Σ Ey = E2 sin 37˚- E1

= (2.4x103)(sin 37˚) - (6.67x103) = - 5225.6 N/C

ER ( . ) ( . )1916 7 522562 2 = 5566 N/C

E1

37º

E2

ER

θ

tan.

.1 52256

1916 7= 70˚

180˚ + 70˚ = 250˚

ER (5566 N/C, 250˚)