electrostatics: electric charges, potentials and fields
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A powerpoint presentation about electrostatics.TRANSCRIPT
ONE:ONE:ELECTROSTATICS
In This ChapterpPart 1: The Electric Charge• To define and explain the concept of charge and charge • To define and explain the concept of charge and charge configurations and its properties
• To utilize Coulomb’s Law to compute for the electric force
Part 2: The Electric Field• To define and explain the concept of electric field, it’s properties and Gauss’s Law
• To compute for the electric field of different charge distributions.To compute for the electric field of different charge distributions.
Part 3: The Electric PotentialPart 3: The Electric Potential• To define and explain the concept of electric potential and relate it to the electric field
• To compute for the electric potential of different charge distributions
PART ONE: CHARGEPART ONE: CHARGE
El t i Ch1. Electric Charge2. Conductors and Insulators
3 The Charging 3. The Charging ProcessC l b’ L4.Coulomb’s Law
The Electric PhenomenonElectric interactions have long been observed.The history of electricity reaches back to the Ancient Greeks when attraction Greeks, when attraction and repulsion are observed when “rubbed amber” are l d placed near common
materials.Indeed the word “electric” Indeed, the word electric comes from the Greek word for amber “elektron”We begin our examination of electricity with ELECTROSTATICS the ELECTROSTATICS – the study of charges at rest
1. Electric ChargeBasic element of ElectricityAn intrinsic property of the An intrinsic property of the fundamental particles that
k ttmake up matter
Two types*:Positive ChargePositive ChargeNegative Charge
N d b B F kliNamed by B. Franklin
1 El t i Ch At i M d l1. Electric Charge: Atomic Model
All matter consists of atomsatomsAtoms are made of a nucleus (neutron and nucleus (neutron and proton) and electron/(s) revolving around the revolving around the nucleus. (Orbital Model)Protons are positively Protons are positively charged and Neutronsare neutrally charged.are neutrally charged.Electrons are negatively chargedcharged.
1. Electric Charge: Atomic ModelParticle Mass Charge
ELECTRON 31 k 6 19 CELECTRON, e‐ 9.11 x 10‐31 kg ‐ 1.6 x 10‐19 CPROTON, p 1.67 x 10‐27 kg + 1.6 x 10‐19 C, p 7 gNEUTRON 1.67 x 10‐27 kg None
**The electron is 2000 less massive as a proton yet The electron is 2000 less massive as a proton, yet they possess the same amount of charge (though opposite in sign)opposite in sign)
1. Electric ChargeAll charges follow the qualitative “fundamental qualitative fundamental law of electrostatics”: + +
“ Like charges repel, unlike ‐ ‐charges attract”
+ ‐
1. Electric ChargeThis is how, the fundamental law of electrostatics was observed:observed:
1. Electric ChargeElectric Charges have two properties:
“Charge is Quantized”All observable charges in nature occur in discrete packets or in integral g p gamounts of the fundamental unit of charge e.Any charge Q occurring in nature can be written Q = + Ne
“Charge is Conserved”When you effect a transfer of charge: If an electron goes from object A to object B, object A becomes positive and object B becomes negative. The net charge of the two objects remains constant; that is, charge is g j ; , gconserved.
E i t i i t ti h h d ti l t d d Even in certain interactions, where charged particles are created and annihilated, the amount of charges that are produced and destroyed is equal, so there is conservation.
1. Electric Charge: Physical Properties
Unit Deri ation ValueUnit
C l b
Derivation
D i d f
Value
• e = 1 60 x 10‐19 C• Coulomb• abbr. C
• Derived from the concept f
• e = 1.60 x 10 19 C
• Type: Derived
of current which is one Derivedof the 7 SI Fundamental Units
Checkpoint 1.1A charge of magnitude 50 nC (1 nC = 10‐9C) can be produced in the laboratory by simply rubbing two laboratory by simply rubbing two objects together.How many electrons must be transferred to produce this charge?transferred to produce this charge?Hint: (Use Q=Ne)
If you rub an inflated balloon If you rub an inflated balloon against your hair, the two materials attract each other, as shown in the Figure. Is the amount of charge Figure. Is the amount of charge present in the system of the balloon and your hair after rubbing
(a) less than (a) less than, (b) the same as,Or (c) more than the amount of charge
t b f bbi ?present before rubbing?
2. Conductors and InsulatorsConductors
are materials where electrons are free to move about the are materials, where electrons are free to move about the entire material (ex. Cu and other metals)
Insulators Insulators are materials, where electrons are bound to a nearby atom, rendering no motion (ex Wood and glass)rendering no motion (ex. Wood and glass)
IonAn atom where electron/(s) is/are added or removedAn atom where electron/(s) is/are added or removed.
N ll d i l i ll l d b l b Normally, a conductor is electrically neutral due to a balance between positive and negative charges. So in order to create a net charge, free electrons are added or removed from the lattice.f
3. The Charging ProcessA macroscopic object can be…
Net Charge Condition Process
Electrically Neutral If p = e‐ NonePositively Charged If p > e‐ Remove electronNegatively Charged If p < e‐ Add electron
Only electrons can be transferred due to the atomic
g y g p
ystructure, and the minimal amount of energy required.Protons are bound by very “strong forces” so their removal is very hard to accomplish.y p
3. The Charging ProcessThe Electroscope
Is a device for Is a device for detecting electric chargesg
The Diverging Leaves:Two gold “leaves” Two gold leaves diverge when a charge is placed near or in pcontact with the bob.
The leaves return to normal, when charges are no longer present in the bob
3. The Charging ProcessThere are two ways of chargingBy Conduction ‐ charging by contact
Implements an effective Implements an effective transfer of electrons
By Induction – charging y g gwithout contact, only by placing objects close to each otherother
Implements only motion of charges within a materialg
But, How do you produce NET h fi ?a NET charge first?
RUB!!!
The Charging Process: By Conduction Case 1
ELECTRON FLOW
The Charging Process: By Conduction Case 2
ELECTRON FLOW
The Charging Process: By Induction Case 1
The Charging Process: By Induction Case 2Ground – A very large conductor that can supply an unlimited amount of charge (such as the earth, g ( ,extremely negatively charged)
Checkpoint 1.2Question1:
When a glass rod is rubbed by silk, which of the two materials i i i hacquire a net positive charge?
Answer 1:A f th t l th th t th it W Any of the two, as long as the other gets the opposite. We can not know for certain which charge is which. We can only arbitrarily assign a charge
Question 2: If lk d ff l f f If you walk across a rug and scuff electrons from your feet, are you negatively or positively charged?
Answer 2:Answer 2:You are positively charge, since electrons were scuffed off/from your feet!
Quiz # 2 (1/4 sheet, 10 points)Two identical spheres are charged by induction and then separated; sphere 1 has charge Q and sphere 2 has then separated; sphere 1 has charge Q and sphere 2 has charge −Q. A third identical sphere is initially uncharged If sphere 3 is touched to sphere 1 and uncharged. If sphere 3 is touched to sphere 1 and separated, then touched to sphere 2 and separated, what is the final charge on each of the three spheres? what is the final charge on each of the three spheres? ‐‐(Answers Q1 = Q/2, Q2 = −Q/4, Q3 = −Q/4) (Answers Q1 Q/2, Q2 Q/4, Q3 Q/4)
4. Coulomb’s LawCoulomb's law, developed in the 1780s by French physicist Charles Augustin de CoulombAugustin de Coulomb.
The magnitude of the electrostatic f b t t i t h i force between two point charges is directly proportional to the magnitudes of each charge and inversely proportional to the inversely proportional to the square of the distance between the charges.
This has the same form as Newton’s Third Law of Motion:The electric force exerted by the two objects on one another have the same magnitude but opposite in directiong pp
4. Coulomb’s Law: FormulaForm:
|q1| and |q2| are the magnitude of the charges|r| is the distance between the charges
kC is the electrostatic constant
ε0 is the permittivity of free space + +
F21
Diagram for force notation
4. Coulomb’s Law:4. Coulomb s Law: Force exerted by a system of chargesIf you need to find the net force exerted on a charge by a group or system of forces, we need to implement vector addition!Because forces are vectors, they superimpose!Because forces are vectors, they superimpose!So F1 = F12 + F13!
+ Q+ Q2
F12
- +Q1 Q3F13
4. Coulomb’s Law: ImplementationThe magnitude of the force is determined by Coulomb’s law.yThe direction of the force is dictated by the fundamental law of electrostaticsIf two or more charges are
i i present, we use superposition principle to calculate the net forceforce.By superposition, we use vector algebra for PHYS 13 vector algebra, for PHYS 13, we utilize unit vector notation.
4. Coulomb’s Law: ExampleConsider three point charges located at the corners of a located at the corners of a right triangle as shown in the Figure, where q1 = q3 = 5.0 μC, g , q1 q3 5 μ ,q2 =2.0 μC, and a = 0.10 m. Find the resultant force
dexerted on q3.
Ans: F3 = (‐1.1i + 7.9j) N
Calculate the net force on q3
q1 5 μC, (0m, 0.6m)q1
‐2 μC (0 8m 0m)
q3
2 μC, (0.8m, 0m)
q2 ‐5 μC, (0m, ‐0.6m)
4. Coulomb’s Law: ExampleTwo identical small charged spheres, each having a mass p gof 3.0 x 10‐2 kg, hang in equilibrium as shown the in Fi Th l th f h Figure. The length of each string is 0.15 m, and the angle θ is 5 0° Find the angle θ is 5.0 . Find the magnitude of the charge on each sphere. The distance pbetween the charged spheres is 0.026m
Ans: |q| =4.4 x 10‐8 C
PART TWO: FIELDPART TWO: FIELD5. Electric Field6. Electric Field Lines
M ti f Ch 7. Motion of Charges in E‐Fields
8. Electric Field of Ch Charge Distributions
9. Gauss’s Law
How do we know?How do we immediately know that there is a force acting on an object?acting on an object?
When we see a change in its When we see a change in its state of motion!
N f h i Now, most of the time, we intuitively think of force as a “contact” interaction between two objects.
I l t t ti h In electrostatics however, there are times that there is no contact but still there is
l i !acceleration!
The Fundamental Law of Electrostatics“Like charges repel, unlike charges attract!”So how is it possible…unlike charges attract!That even without contact there are forces Repulsion and attraction can easily be observed
contact, there are forces between the two masses, evident by the repulsive through a visual change in the distance between
evident by the repulsive and attractive nature of interaction?two particles!interaction?
This is due to the concept Even without contact!This is due to the concept of ELECTRIC FIELDS!
5. Electric FieldsAre said to exist in regions surrounding a charged object ( h )
Force(source charge)!
When another charge (test gcharge) enters this region, this test hcharge experiences a force!
5. Electric Field
The Electric Field vector E, at a point vector E, at a point in space is defined as the ratio of the
E => Vector Field Travels at the speed of lightas the ratio of the
electric force experience by a test D ll experience by a test charge to the magnitude of the
Acts on tests not on its own source
Does not actually require test charges to magnitude of the
test charge!
gcompute E
Computing for the Electric FieldThis version will only work on point charges and systems of point charges!
When the force and test h lcharge values are given:
When only the source When only the source charge value is given:
*The unit of E is N/C!
6. Electric Field LinesWe can visually represent electric fields as field lines.
Caution: Electric Field lines are just imaginary – a mere
i f h representation of the electric vector field!
We note that the electric fi ld li i di t th field lines indicate the direction to which the force will be exerted by a positive will be exerted by a positive test charge!
How to Draw Electric Field Lines?
• Begin/End• Symmetry1 • Number
• Density2 • When Far• Star‐Crossed3
Try drawing electric field lines on the following charge configurations:
q
+4q ‐3q
q‐5q
+4q ‐3qqq 2q
7. Motion of Charges in Electric FieldsWhen a test charge ventures in an electric field, ventures in an electric field, it experiences a force q0E!It will accelerate It will accelerate following Newton’s Second Law withSecond Law with
An electron is projected into a uniform horizontal electric field (E = 1000 N/C, i) ( 6 )with a horizontal velocity (v0= 2 x 106 m/s, i) in the direction of the field. How far
does the electron travel before it is brought momentarily to rest?
2‐D Motion in FieldsAn electron enters a region of uniform electric field of E = 200 j N/C and zero‐g, with an initial velocity of 3 x 106 i m/s. F h i ht f f th
‐‐From a height of 10m from the ground. Find the following:
10 m
1. Acceleration in the field. Th ki d f i
H = 1
2. The kind of motion experienced by the electron in the fieldthe field.3. The range of the electron in the field!
Find the RANGEthe field!
Quiz # 3 (1/2 sheet, 10 points)Suppose a charged volleyball (Q = ‐16C, m = 4kg) was launched with a velocity of 3 x 102 m/s at an angle of y 3 / g30o with respect with the horizontal.
Find the volleyball’s “Maximum Height”,
If the planet’s gravity field intensity is g = ‐9.8 m/s2 j, d th l t’ l t i fi ld i E N/C jand the planet’s electric field is E = 1.5 N/C j
Hint: Accelerations are vectors they add up!Hint: Accelerations are vectors, they add up!‐Ans: 712.03 mns: 7 .03 m
8. Electric Field Due to Charge Distributions
Two Kinds of Charge Distributions:
Discrete1. Electric Dipole2. Systems of Point Chargesy g
C tiContinuous1. Linear: Line and Ring Charges2. Surface: Disk and Plane Charges3. Volume: Spheres and Cylinders3 p y
a. Electric DipoleElectric dipoles are systems composed of systems composed of two equal and opposite charges q, separated by g q, p ya small distance L.Electric dipole Electric dipole moment, p describes the strength and e s e g a dorientation of electric dipoles.pp, points from the negative charge to the negative charge to the positive charge
Example: Electric DipoleA molecule of water vapor causes an electric field in the surrounding space as if it were an electric dipole Its surrounding space as if it were an electric dipole. Its dipole moment has a magnitude p = 6.2 x 10‐30 C•m.
Wh t i th it d f th l t i fi ld t di t What is the magnitude of the electric field at a distance z = 1.1 nm from the molecule on its dipole axis?
Ans: p = 8.4 x 107 N/CAns: p 8.4 x 10 N/C
b. System of Point ChargesJust as Forces are vectors, Fields are also vectors Fields are also vectors hence they also follow the superposition the superposition principle, ie
The net field at a certain point is just the vector sum of the individual contribution of each point chargesp g
Example/Assignment: System of Point Charges
Find the electric field caused by an electric caused by an electric quadrupole (group of point charges), as
B = (2m, 4m)p g ),shown in the figure to the right at point B.g p
q = 2 5 nC
(‐2m, 0) (2m, 0)
q = 2.5 nC
A
2q‐q ‐q
Ans. (0.5 i + 0.09 j) N/C
For Continuous Charge Distributions
Just like mass corresponds to its effect in density, charge can be found to exist in three spatial forms charge can be found to exist in three spatial forms, with corresponding densities:
Name Symbol SI UnitCharge q CC a ge q C
Linear charge density λ C/m
Surface charge density σ C/m2
V l h d i C/Volume charge density ρ C/m3
c Linear Charges: Lines and Ringsc. Linear Charges: Lines and RingsWe have three types of linear charges:
y is the distance from z is the distance from the center of y is the distance from
centerL is the length of the line charge
from the center of the ringR is the radius of the ringline charge the ring
Each of the field equations are obtained through Each of the field equations are obtained through rigorous mathematical analysis
Example: Lines and Rings1. A rod 14.0 cm long has a uniform linear charge density of λ=‐24μC/m. Determine the magnitude and direction of the 4μ gelectric field along the axis of the rod at a point 36.0 cm from its center.
2. Suppose the rod is extended to infinite length, calculate h i d d di i f h l i fi ld i the magnitude and direction of the electric field at a point 36.0 cm directly above any point along the line.
(ASSIGNMENT # 2) 3. Suppose the rod in the example # 1 i l d i t i l ith t th i i fi d th is curled into a circle, with center on the origin, find the electric field along the central axis of the rod, 36.0 cm above the centerabove the center.
d. Surface Charges: Disks and PlaneWe only consider circular disks and its infinite extension: the infinite planeextension: the infinite plane
R is the radius of the Field is measured l h ldisk
Z is the distance from the center of the disk
normal to the plane
the center of the disk
Example: Disks and Planes1. A uniformly charged disk of radius 35.0 cm carries charge with a density of 7 90 x 10‐9 C/m2 Calculate the charge with a density of 7.90 x 10 9 C/m . Calculate the electric field on the axis of the disk at (a) 5.00 cm, (b) 10 0 cm (c) 50 0 cm and (d) 200 cm from the center of 10.0 cm, (c) 50.0 cm, and (d) 200 cm from the center of the disk. Compare the results.
2. Suppose the charged disk is extended to infinite pp gradius. Calculate the electric field on the axis of the disk at (a) 5.00 cm, (b) 10.0 cm, (c) 50.0 cm, and (d) ( ) 5 , ( ) , ( ) 5 , ( )200 cm from the center of the disk. Compare the results.
9. Gauss’ LawIn our discussion of Gauss’s Law we explore:
1. The Electric Flux2. Gauss’s Law Statement3 Primitive Use of Gauss’s Law3. Primitive Use of Gauss s Law4. Advance Use of Gauss’s Law
i di h fi ld f l Ch Di ib i e. Finding the field of Volume Charge Distributions: Solid Spheres and;
f. Spherical Shells5. Gauss’s Law and Conductors5
9.1 Gauss’ Law: Electric FluxElectric Flux is a quantity that q ymeasures the amount of field lines • Electric Flux
• Unit: Nm2/C• Electric Flux• Unit: Nm2/C
crossing an imaginary surface!
ΦE• Unit: Nm2/C• Unit: Nm2/C
Operationally, the E• Electric Field• N/C• Electric Field• N/COperationally, the
flux is just the dot product of the
E
• Area vector, m2• Area vector, m2pelectric field to the surface area vector. A
Area vector, m• Magnitude: Area of surface• Direction: perpendicularly away from surface
Area vector, m• Magnitude: Area of surface• Direction: perpendicularly away from surface
9 1 Gauss’ Law: Electric Flux9.1 Gauss Law: Electric FluxIf the electric field in a region has a magnitude of 2.0x103 N/C directed to ards the right as sho n towards the right as shown in the figure, what is the value of the electric flux value of the electric flux passing through a rectangular Gaussian gsurface of cross sectional area 0.0314 m2?
If the surface is inclined i h l f i h with an angle of 50o with
respect to the horizontal, find the electric flux!find the electric flux!
9.1 Gauss’ Law: Flux through CubesAn imaginary cube (side length of 3m) is exposed to an electric of 3m) is exposed to an electric field of E = 2i – 4j + k (N/C). Find the electric flux through Find the electric flux through each side of the cube.
i
Assume that one of the vertices
Imaginary cubical surface
is fixed at the origin!E, drawn only a handful just handful just for pictorial representation
9.1 Gauss’ Law: Notes on Flux1. When E and A are
perpendicular, no E th h th
E
passes through the surface, hence no flux.
A
2. When E enters a surface, E and A are
ll l h A
parallel, we have a negative flux
3 When E leaves a
E
3. When E leaves a surface, E and A are anti‐parallel, we h i i Ahave a positive flux E
A
9.2 Gauss’ Law: StatementThe net flux through any Gaussian surface equals the Gaussian surface equals the charge enclosed over permittivity of free space permittivity of free‐space (ε0)
Validity
• Gauss’s Law is valid for
Use
• To find the charge that any kind of charge distribution
• But it is best used for
gcauses a certain field
• To find E, if you know the charge distributions• But it is best used for
charge configurations with hi‐degree of
charge distributions
symmetry
9.3 Gauss’ Law: Primitive Use hIn the primitive use, we
utilize the relation:
Example:p• Four closed surfaces, S1 through S4, together with g 4, gthe charges ‐2Q , +Q , and ‐Q are sketched.
• (The colored lines are the intersections of the surfaces with the page.)
• Find the electric flux through each surface.
9.4 Gauss’ Law: Advanced Use In Advanced Use: we utilize completely the h l d fi iti d hi hwhole definition and high‐
degree of symmetry Gaussian Surfaces:
Gaussian Surfaces:Any hypothetical closed surfaceC b h b t th Can be any shape, but the most useful ones are those that mimics the shape and
f h bl symmetry of the problem at hand.
9.4 Gauss’ Law: Point Charge and Line ChargesSee step by step how we apply AGauss’ Law for Point and Line Ch
EA
Charges:
1. Choose Gaussian Surface A
2. Draw field lines, and vector A
3. Apply relation4 Obtain E4. Obtain E
9.4 Gauss’ Law: Infinite Line Charge1. Choose Gaussian Gaussian Surface D fi ld
y2. Draw field lines, and
Avector A3. Apply pp yrelation
4. Obtain E4. Obtain E
e. Outside: Solid SpheresFor solid spheres, the electric field Solid Sphere the electric field relation varies from inside and outside
Solid Sphere (radius R)
inside and outside the sphere
Gaussian surface (radius r)
id S lid S he. Inside Solid Spheres
For solid spheres, the electric field G i f the electric field relation varies from inside and outside
Gaussian surface (radius r)
inside and outside the sphere
Solid Sphere (radius R)
f. Spherical Shells Spherical Sh ll( di R)
Inside the shell, the field is zero why?
Shell(radius R)
is zero, why?
Gaussian surface (radius r)
O t id th h ll th Outside the shell, the field is:
9.4 Examples for Gauss’s Law1. A solid sphere (R = 10 cm) carries a uniform charge density of ρ =2 8 μC/m3 calculate the electric field at density of ρ =2.8 μC/m3, calculate the electric field at the following r: 5 cm, 12 cm, 100 cm from the center of the spherethe sphere.
2. A spherical shell of charge (R = 10 cm) carries a uniform charge density of σ = ‐5nC/m2, calculate the g yelectric field at the following r: 5cm, 12 cm, 100cm from the center of the shell.
9 5 G ’ L d C d t9.5 Gauss’ Law and ConductorsConductors have free charges that Conductors have free charges that are able to move around the conductor.
If there is an electric field inside the conductor, there will be a net ,force on this charges causing a momentary electric current.
However, unless there is a source of energy to maintain this current, the h ill l di ib charges will merely redistribute itself to nullify the field created inside the conductor
This is known as Electrostatic E ilib i ! Equilibrium!
9.5 Gauss’ Law and ConductorsA conductor in Electrostatic Equilibrium has the following properties: Which of the following properties:
1. The electric field is zero
Which of the following should you use as refuge during a thunderstorm?
1. The electric field is zero everywhere inside the conductor.2 Any net charge on the
a. Underneath a treeb. Open field
L l i 2. Any net charge on the conductor resides entirely on its surface.
c. Lowest place in townd. Inside a car
3. The electric field just outside the conductor is perpendicular to its surface perpendicular to its surface and has a magnitude σ/ε0, where σ is the surface charge density at that pointdensity at that point.
PART THREE: POTENTIALPART THREE: POTENTIAL
10 Electric Potential and 10. Electric Potential and Potential Difference
11 Potential of Charge 11. Potential of Charge Distributions
12 Calculating E from V 12. Calculating E from V and vice‐versa
13 Equipotential Surfaces13. Equipotential Surfaces
10 Electric Potential10. Electric PotentialFrom Physics 3:
graviFrom Physics 3:Suppose you lift a book of mass m to a height h…
ity
g
What is the work you have done?
WAF = mghh A
ppl
What is the work done by the gravity field?
lied forcg y
Wg= ‐mgh
e
10 Electric Potential10. Electric PotentialFrom Physics 3:
U = mghFrom Physics 3:
Suppose you lift a book of mass m to a height h… gr
How does the potential energy change as the book is lifted?
ravity
g
It increasesh
How does the potential energy change as the book is lowered to h d?the ground?
It decreases U ZIt decreases U = Zero
10. Electric PotentialAs we all know, gravity is a What does g yconservative force, and that is why th i t ti l
ΔU= ‐W mean?there is a potential energy associated with it!
• It means that as the potential
• It means that as the potential with it!
We can generally
the potential energy increases, the work that
the potential energy increases, the work that We can generally
say thatthe work that must be done is
ti
the work that must be done is
ti ΔU = ‐W
negative or against the field!negative or against the field!
10 Electric Potential10. Electric PotentialNow for the analogy:Now for the analogy:
Suppose we place a test charge in a certain field +in a certain field…
It will possess potential energy
+EIt will possess potential energy
because of its position in the field.
Efield.
The ratio of the potential The ratio of the potential energy to the test charge is known as Electric Potential, V,
10 General Expression for V10. General Expression for VElectric Potential Equation SetElectric Potential Equation Set
Potential Energy‐Work Relation Relation
E force from E fieldE‐force from E‐field
El t i W k Electric Work
f l lV from Electric Potential
10. Electric Potential VTherefore, the Electric Potential is:Potential is:
Unit
• Volts
Defined
• Defined as the d b
Properties
• Scalar Quantity
To Calculate
• No need the value f h h • After A. Volta
• 1 V = 1 J/C• 1 V = 1 Nm/C
energy stored by a test charge since it is in a field.
• V is the scalar property of the field
of the test charge to calculate value since we only need to know the need to know the magnitude of the field and point of measurement
10. Electric Potential and the Electric FieldSince electric field lines emerge from ELECTRIC FIELDgpositive charges and end on negative h
High V Low Vcharges
l i fi ld li +q0
Electric field lines point to a place of lower potential
As a positive charge moves in the di i f h fi ld i i l lower potential
C fi ith
direction of the field, its potential decreases, the electric force is directed to the right
Confirm with Fundamental Law of Electrostatics
As a positive charge moves against the field its potential increases the Electrostatics field, its potential increases, the electric force is directed to the right
10. Electric Potential CheckpointSuppose we have a negative test charge Bnegative test charge smiley.
B
Find the change in A Cgpotential and the direction of the force
A C
d ect o o t e o cein the following scenariosscenarios
ELECTRIC FIELD
High V Low Vg
10. Potential DifferenceThe potential difference ΔV is the difference of
t ti l b t t potential between two points A and B in an Electric Field!CheckpointIn figure, points A and B are in an electric g , p
field.
I th t ti l diff V V Is the potential difference VB – VA, positive, negative, or zero?
Suppose a negative charge is placed at A and then moved to B, the change in
t ti l i iti ti ?potential is: positive, negative or zero?
11. V of Charge DistributionsIn this section we will consider the calculation of the following charge distributions:following charge distributions:
Discrete ChargesPoint ChargegSystem of Point Charges
Continuous Charge DistributionsContinuous Charge DistributionsLines and Ringsi k d lDisks and Planes
Sphere
11a: V of Discrete ChargesSingle Point Charge System of Point ChargesSingle Point Charge System of Point Charges
q rq1 r1 q3q
P
rPr2 r3
q2
*this is known as the Coulomb Potential
Notes: 1 We implement sign rules for charges here since V is a scalar 1. We implement sign rules for charges here since V is a scalar
quantity2. We only need to know field/potential point P and its distance r
from the charge to compute for V
11a: Discrete Charges Example1. What is the electric potential at a distance r = 0 529x10‐10 m from a proton? (This is the average 0.529x10 m from a proton? (This is the average distance between a proton an electron in a H atom)Wh t i th t ti l f th l t d th What is the potential energy of the electron and the proton at this separation?
2. Four (‐1C) charges are arranged in each corner of a 2. Four ( 1C) charges are arranged in each corner of a square of side 1m. Find the potential at the center of the square.the square.
11b: V of Lines and Rings11b: V of Lines and RingsInfinite Lines Infinite RingsInfinite Lines Infinite Rings
The potential V is given by: The potential V is given by:
N h V Note that V = 0 at a.Note that R is radius, x is di t f th t f distance from the center of the ringr
x
R
11b: Example for V of Lines and Rings1. Compute the potential at a perpendicular distance (r = 20 5 cm) from an infinite line (λ=26 3 nC/m) if (r = 20.5 cm) from an infinite line (λ=26.3 nC/m) if the potential is found to be zero at a= 10.5cm from the lineline.
2. Calculate the potential due to a ring (total charge 0.5mC, radius 20cm) at a point 25cm from the center pof the ring.
11 V f Di k d Pl11c: V of Disks and Planes
Di k I fi i PlDisks Infinite PlaneThe potential V is given by: The potential V is given by:g
Note that R is the radius, x is the distance from the center
Note that Vo is the potential of the disk, |x| is the absolute di t f th lof the disk. distance from the plane.
x |x|
RR
11c: V of Disks and Planes1. Calculate the potential of a disk at 36 cm from the center of the disk if the disk carries (σ=‐7 μC/m2) and center of the disk if the disk carries (σ=‐7 μC/m ) and has a radius of 5cm.
2. An infinite plane of charge has a surface charge density of 3.5μC/m2. How far apart are the “equipotential” surfaces whose potential differ by q p p y100V? Note: Equipotential means “the same” potential.
11d: V of Spherical Shells11d: V of Spherical ShellsInside the Shell Outside the ShellInside the Shell Outside the Shell
The potential V inside the shell is constant and is given
The potential V outside the shell varies with the distance shell is constant and is given
by:shell varies with the distance r from the center.
R is the radius of the shellR is the radius of the shell.
Rr
11d: Example for V of Spherical ShellsUnlike the Electric Field E, V is continuous over a region of space. This is illustrated in the discontinuity g p yin the electric field fluxing through an infinite plane.
The continuity of V is illustrated here”
Calculate the potential due to a spherical shell of charge f di d i t t l h f C t of radius 10 cm and carrying a total charge of 50 μC at
the following points:) b) ) d) a) 0cm b) 5 cm c) 9 cm d) 11 cme) 15 cm f) 50 cm
12. Calculating E from V and vice‐versaLet us remind ourselves with
E can be found E can be found V can be found V can be found E can be found from a change in potential
E can be found from a change in potential
V can be found by summing changes in E
V can be found by summing changes in E
f hNow, for the zeroes:If E is zero, what is V? Constant!,If V is zero, what is E? ZERO!
13. Equipotential Lines and SurfacesEquipotential Line
are lines drawn in an are lines drawn in an electric field such that that all the points on the line all the points on the line are at the same potential.
Equipotential SurfaceEquipotential Surfaceis a surface, all points of which are at the same which are at the same potential.
13. Equipotential Lines and SurfacesEquipotential Lines and Surfaces are always perpendicular to the electric field lines!perpendicular to the electric field lines!
13. Equipotential Lines and SurfacesMovement along an Equipotential line requires no work because such movement is always perpendicular work because such movement is always perpendicular to the electric field.
13. Metals and the EquipotentialHow much work would it take to drag over the gsurface of a conducting metal, a positive charge q from Point A to Point B?
Answer is NONE!
Because metals are equipotential volumes and equipotential volumes and they have equipotential surfaces See Gauss’s Law surfaces. See Gauss s Law for explanation!