W. Sautter 2007

Electrostatics is the study of the effects of stationary charges on each other in their surroundings.

Charges are created by the transfer of electrons to or from one body to another. (Protons are NEVER transferred.)

Objects with equal numbers of protons and electrons are neutral. They have no net charge.

Objects with more electrons than protons are charged negatively

Objects with less electrons than protons are charged positively.

- +

+ +

- +

--

++

- -Likesrepel

Likesrepel

Unlikesattract

Unlikesattract

6.25 x 10 18electrons1 coulomb

Charge on 1 electron = - 1.6 x 10 –19 coulombs

Charge on 1 proton = + 1.6 x 10 –19 coulombs

Fg = gravity force betweenm1 and m2 separated by

a distance r

G is the Universal Gravitational Constant

The weight of an object is its mass times g’, the

gravity value at location r

Fe = electrostatic force Between m1 and m2

separated by a distance r

k is Coulomb’s Constant

Coulomb’s Law is an inverse square lawsimilar to the Law of Gravitation

It is dissimilar in that electrostatic forces can be attraction or repulsion. Gravity is attraction only.

Electrostatic force is strong, gravity is very weak.

Gravity is classified as a weak force because huge amounts of mass are required to create a reasonably

large force.Note the very small value of the force constant.

Electric forces is classified as a strong force because small charge quantities can to create large forces.

Note the very large value of the force constant

Electric charges are detected by the presence of an electric field (E).

Recall that a gravity field can be detected by its influence (attractive force) on a mass (often called a test mass).

Electric fields are detected by their influence (attractive or repulsive forces) on a charge (often called a test charge).

Electrically charged bodies can be created by physically rubbing electrons off one object on to another. For example, rubbing a rubber rod with fur will transfer electrons from the fur to the rod which becomes negatively charged (it has extra electrons).

Rubbing a glass rod with silk will transfer electrons from the rod to the silk. The rod is charged positively (it has missing electrons).

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--

-

--

--

-

- - -- - -

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Charged rod

Leaves of electroscopeDiverge (like charges repel)

An Electroscope is an instrument used to measure

the presence of an Electric field (presence of charged bodies).

Moveableleaves

- - ---- --

NEUTRALOBJECT

- - ---- --

CHARGING BY CONDUCTION

WHEN OBJECT IS TOUCHEDBY THE CHARGED ROD

ELECTRONS MOVE FROMTHE ROD TO THE SPHEREUNTIL ELECTROSTATIC

EQUILIBRIUM IS REACHED

OBJECT IS NOW

NEGATIVELYCHARGED

- --

-

- -

INDUCTIVE CHARGING

CHARGED RUBBER ROD

(EXCESS ELECTRONS)

Electrons on sphere move to the opposite side due to repulsion of electrons on the charged rod

Electric Field

E is always out of plus (+) into minus (-)

Charged Plates

Recall that a gravity field (g) is measured by dividing the force acting on it (its weight(w) in Newtons ) by the mass quantity (m) in kilograms

An electric field (E) is measured by dividing the force acting on it (in Newtons ) by the charge quantity (q) in coulombs (C ).

g in N / Kg

E in N / C

+

----

+

----

EE

ScaleScale

As Electric field strengthincreases, the Force on a test

Charge increases.This is similar tothe weight of a

Mass increasingin a strongerGravity field

Test chargeTest charge

Electric Field

Charged Plates

+ Force

E = Force (N) / Charge (Coul)

+

Lines of Flux spread over a greater area as distance from charge increases and field strength weakensE = k q / r2

(1) E = F/q (by definition)(2) F = kq1 q2 / r2 (Coulomb’s Law) (3) E = kq1 q2 / r2 q2 ( by substitution)

E = kq / r2

q = point charge in coulombsk = coulomb’s constant r= distance from charge in

metersE = electric field strength

at that point (N / C)

Electrical Potential is defined as the work required to move a charge over a distance in an electric field.

Electrical potential is measured in volts.

One volt equals one joule (work) divided by charge (coulombs)

In a uniform electric field (a field between two parallel charged metal plates) , potential (V) equals work (W) divided by charge (q).

Therefore, since work equal force times distance and force equals the electric field strength (N/C) times charge in coulombs (C) , potential equals electric field times distance the charge moves.

V = W / q = (F x d) / q = (F / q) x d = E x d

V (volts) = E ( N/C) x d (m)

Electric Field

Charged Plates

x Point A

x Point B

+ work+ work

V = Work (Joules) / Charge (coul)

In an electric field issued from a point charge the field varies inversely with the square of the distance from the charge, therefore the work required to move the charge is continuously changing.

This is similar to the varying gravity field surrounding a mass. The work required to move a mass in that field also continuously changing.

To find the work required to move the charge in a varying field we must integrate the force times distance equation using calculus.

The following frames show the use of calculus to find work form force versus distance relationships. If your course does not require calculus, skip these frames

FORCE

(N)

DISPLACEMENT (M)

X1 X2

WORK = AREA UNDER THE CURVEW = F X (SUM OF THE BOXES)

WIDTH OF EACH BOX = X

AREA MISSED - INCREASINGTHE NUMBER BOXES WILL

REDUCE THIS ERROR!

AS THE NUMBER OF BOXESINCREASES, THE ERROR

DECREASES!

BOX METHOD

Finding Area Under Curves Mathematically

• Areas under force versus distance graphs (work) can be found mathematically. The process requires that the equation for the graph be known and integral calculus be used.

• Recall that integration is also referred to as finding the antiderivative of a function.

• The next slide reviews the steps in finding the integral of the basic function, y = kxn.

INTEGRATION – THE ANTIDERIVATIVEINTEGRATION IS THE REVERSE PROCESS OF

FINDING THE DERIVATIVE. IT CAN ALSO BE USEDTO FIND THE AREA UNDER A CURVE.

THE GENERAL FORMAT FOR FINDING THE INTEGRAL OF A SIMPLE POWER RELATIONSHIP, Y = KXn

ADD ONE TO THE POWER

DIVIDE THEEQUATION

BY THE N + 1

ADD A CONSTANT

is the symbolfor integration

APPLYING THE INTEGRAL FORMULA

GIVEN THEEQUATION

FORMAT TO FIND THE INTEGRAL

Integration can be used to find area under a curve betweentwo points. Also, if the original equation is a derivate, then

the equation from which the derivate came can be determined.

APPLYING THE INTEGRAL FORMULAFind the area between x = 2 and x = 5 for the equation y = 5X3.First find the integral of the equation as shown on the previous

frame. The integral was found to be 5/4 X4 + C.

The values 5 and 2 arecalled the limits.

each of the limits isplaced in the integratedequation and the resultsof each calculation aresubtracted (lower limit

from upper limit)

W= Fdr = ( k q1 q2 / r2 )dr00

r

00

r

W = k q1 q2 r-1 = - k q1 q2 / r |00

r

W = - k q1 q2 / r

( 1/ infinity ~ 0 ) therefore

From Infinity ( a great distance) to a point in the field

Coulomb’s Law

Work = - G m1 m2

r

Recall: Work to move a charge in a gravity field:

By Analogy: Work to move a charge in an electric field:

Work = - k q1 q2

r

q

Dividing the work equation obtained in the previous slides by the charge quantity of the point charge we get a

relationship which describes absolute potential at point in the field

Absolute potential is measured by using theWork to move a charge from infinity to a point

+

Work to move a charge from infinity to a point in an electric field is called Absolute Potential

1 volt = 1 joule/ 1coulV = k q / r

Point Ax + work+ work

Potential = Work / Charge

Volts = Joules / Coulomb

Recall from previous slide:

W = - k q1 q2 / r

VAbsolute = ( - k q1 q2 / r ) / q2

VAbsolute = - k q1 / r q1 = point charge

r= distance from the point charge

Absolute potential is measured from infinity to a point r in the field

W = - k q1 q2

rV =

q / q1\

\= k q / r

V = k q / r

Where V is the absolute potential at a point (r) in the electricfield caused by charge (q) , we can now find the potential

difference between two points in the field.

Absoluteat point A

Absoluteat point B

Potential DifferenceFrom point A to B

Capacitors are electrical devices used to store electrical energy.They are not to be confused with batteries which create

electrical energy via chemical reaction.

The structure of a capacitor is shown on the next slide.Essentially, electrons are pumped onto one of the metal

plates shown and pushed off the other plate by the electricfield developed on the first plate. The electrons are pumped

be a battery, power supply or other voltage source.

Once the charge is isolated on the metal plates it representsstored energy since when the capacitor is connected to anelectrical circuit, electrons will flow from the plate of high

electron concentration through the connecting device, thus releasing energy and back to the plate of low electron

concentration until charge equilibrium is reached.

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C (farads) = q (coul) / V (volts)

1 F = 1 coul / 1 volt

Large capacitor can hold more charge at lower voltages and

can store more energyCapacitors

Store electricalenergy

A Farad is a unit of capacitance

Electronsin

Although it may seem illogical, inconventional current the + plate

has the excess electrons !

(1) Large capacitors can hold more charge andlarger amounts of energy than smaller ones.

CAPACITOR TRENDS

(2) The closer the plates are together, the more energy that can be stored since the electric field

between the plates becomes stronger.

(3) The larger the dielectric constant the lessthe tendency of electrons to “jump the gap”

(short out) between the plates at higher voltages and more energy can be stored.

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Dielectric (non conductive material)

Metal plates (A) in meters2

Plate separation (d) in meters

C = e0k A / d

e0 = permittivity constant8.85 x 10-12 Coul2 / N m2

As plate area increases, C increasesAs plate separation increases, C decreases

K = dielectric constant (air = 1.0)It depends on the nature of the dielectric

Charging a Capacitor

CHARGE

(C)

VOLTAGE (V)

V

C

Area under the graphgives the work to charge

the capacitor

The slope of the graphline equals the capacitance (C )

C = q / V

W = ½ qV, Also since q = CV, by

SubstitutionW = ½ CV2, and since V = q / C

W = ½ q 2/ V

C(total) = C (1) + C (2) + C (3)

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Effectively, the capacitors in parallelbecome one larger capacitor

Since each individual capacitor is directly connected to the voltage source

(battery) each bears the same voltage

The charge on each capicator is dependenton its individual capacitance. (Plate

size, separation, dielectric)

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1/C(total) = 1/ C (1) + 1/ C (2) + 1/C (3)

Capacitors in series have different characteristics thanThose connnected in parallel

(1) Since they are not all directly connnected to the voltage source, the voltage on each varies according

to its individual capacitance. Larger ones havelower voltage than smaller ones.

(2) Each capacitor, despite its size, has the same charge.

1/C(total) = 1/ C (1) + 1/ C (2) + 1/C (3)