1

Physics STPM

12. Electrostatics 12.1 Coulomb’s law 12.2 Electric field 12.3 Gauss’ law 12.4 Electrical potential OBJECTIVES

(a) state Coulomb’s law and use the formula F = Qq / 4πε0r

2

(b) explain the meaning of electric field, and sketch the field pattern for an isolated point charge, an electric dipole and a uniformly charged surface

(c) define the electric field strength, and use the formula E=F/q

(d) describe the motion of a point charge in a uniform electric field

(e) state Gauss’s law, and apply it to derive the electric field strength for an isolated point charge, an isolated charged conducting sphere and a uniformly charged plate

(f) define electric potential (g) use the formula V = Q / 4πε0r

2

(h) explain the meaning of equipotential surfaces (i) use the relationship E = -dV/dr (j) use the formula U = qV

12.1 Coulomb’s Law a) What is Electric Charge? - An intrinsic property of

protons and electrons, which make up all matter, is electric charge.

- A proton has a positive charge, and an electron has a negative charge.

b) Properties of electric charge - Two types of electric charge, positive and negative; a

proton has a positive charge, and an electron has a negative charge.

- The SI unit for measuring the magnitude of electric charge is the coulomb (C).

- The electric charge is said to be quantized. The smallest amount of free charge is e=1.6×10

-19 C. Any electric

charge, q, occurs as integer multiples of the elementary charge e, Q = ne

- Two electrically charged objects exert a force on one another, called as electrostatic force: like charges repel and unlike charges attract each other.

c) The strength of electrical force depends on

- The distance between charges - The amount of charge on each object d) Charles Augustin Coulomb (1736 – 1806) - Studied electrostatics and magnetism - Investigated strengths of materials - Identified forces acting on beams (1785) e) Coulomb's Law states that:

- The electrostatic force between two charged objects is proportional to the quantity of each charge and inversely proportional to the square of the distance between charges.

221

4 r

qqF

o (Newtons)

where: q = charge, measured in

Coulombs (C). k = Coulombs constant = 8.99 x 10

9 N.m

2/C

2.

- Or 2

21

4 r

qqF

o , where

k = 4

1

0 2

29

C

Nm 1099.8 ,

ε0 = permittivity constant = 2

212

Nm

C1085.8

Opposite charges: F is attractive (-) Like charges: F is repulsive (+)

f) Difference and Similarities between Electricity and Gravity

- Mathematical form of the Coulomb law and law of gravitation very similar

221

r

QQkFe

- gravitation is always attractive (no negative mass)

- electrical force can be both attractive or repulsive

- Electric force is dominant in the atomic world

- Gravitational forces dominates on the macroscopic scale: people, planets, galaxies

- Electric forces are more stronger !

221

r

mmGFg

2

29109

C

Nmk

2

211107.6

kg

NmG

Example 1: What is the electrostatic force between two positive

charges? [q1 = 4 μC, q2 = 8 μC, r = 10 cm]

F = kqq/r

2 = (9 x 10

9) (4 x 10

-6) (8 x 10

-6) / (0.1)

2 = 28.8 N

- If ‘r’ is doubled, the force is reduced by a factor of 4

g) The force on a point charge due to two or more other point charges

- There are three charges q1, q2 and q3. What would be the net force on q1 due to both q2 and q3? - First, find the magnitude and direction of the force

exerted on q1 by q2 (ignoring q3). - Then, determine the force exerted on q1 by q3 (ignoring

q2). - The net force on q1 is the vector sum of these forces.

- If we have n charged particles, they interact independently in pairs, and the force on any one of them, let us say particle 1, is given by the vector sum F1,net = F12 + F13 + F14 etc.

q1 q2

F

r F

2 12. Electrostatics Example 2: Forces F1 and F2 act independently on test charge (q0). [q1 = 2 μC, q2 = 5 μC, q0 = 4 cm]

F1 = kqq/r

2 = 7.2 N (to right)

F2 = kqq/r2 = 2.9 N (to left)

So, Fnet = F1 – F2 = 7.2 – 2.9 = 4.3 N (to right) Note, the larger effect of F1 (even though the charge was

smaller) is due to its closer proximity to q0. Example 3: Three point charges are arranged as shown in the figure

below. (Take q1 = 5.46 nC, q2 = 4.95 nC, and q3 = -2.97 nC.)

a) Find the magnitude of the electric force on the particle at

the origin. F=kq1q2 / r

2 (k=9 x 10

9, q=charge of particle, r=distance

between particles) Force due to q1: F1=(9 x 10

9)(5.46 x 10

-9)(4.95 x 10

-9) / (0.3)

2

F1 = 2.7027 x 10-6

N to the left since repulsive force Force due to q3: F2=(9 x 10

9)(4.95 x 10

-9)(2.97 x 10

-9) / (0.1)

2

F2 = 1.323 x 10-5

N downwards since attractive force To find magnitude, use pythagorean theorem

F2 = (2.7027 x 10

-6)

2 + (1.323 x 10

-5)

2

F = 1.35 x 10-5

N

a) Find the direction of the electric force on the particle at the origin.

To find direction, use

trigonometry

tan = 1.323 x 10-5

/ 2.7027 x 10

-6

= 78.454 degrees from the negative x-axis 180 + 78.454 = 258.454 degrees from positive x-axis

12.2 Electric Fields a) Field Theory - The electric force is not “action at a distance” but is the

action of a field. - A field is a physical entity that extends throughout a

volume of space and exerts forces. - Electric field = E(x,t) and Magnetic field = B(x,t) - Electric fields surround every electric

charge and exerts a force that causes electric charges to be attracted or repelled.

- Force is a push or pull - What would happen to

a (-) charge in each field?

- A charge creates an electric field around itself and the other charge feels that field.

- Test charge: point object with a very small positive charge so that it does not modify the original field

- Electric field at a given point in space: place a positive test charge q at the point and measure the electrostatic

force that acts on the test charge; thenq

FE

- A field is not just an abstract concept that we use to describe forces. The field is real.

- The electric field extends throughout space and exerts forces on charged particles.

- If we place a positive point charge in an electric field, there will be a vector force on that charge in the direction of the electric field

- The magnitude of the force depends on the strength of the electric field.

b) Precise Definition of Electric Field: - We define the electric field in terms of the force it exerts

on a positive point charge - Unit of the electric field: N/C (newtons per coulomb)

- We can then write )(xEqF

- Note that the electric force is parallel to the electric field and is proportional to the charge

- The force on a negative charge will be in the opposite direction

- What is the field created by a point charge q? - Consider a “test charge” q0 at point x.

- Force on q0: rr

qqkF ˆ

20

- Electric field at x: rr

qk

q

FxE ˆ)(

20

c) Superposition of Electric Fields: - Suppose we have many charges. - The electric field at any point in space will have

contributions from all the charges. - The electric field at any point in space is the superposition

of the electric field from n charges is E = E1 + E2 + E3 + E4 + … + En (vectors sum!)

- Note that the superposition applies to each component of the field (x, y, z).

d) Electric Field Lines - We can represent the

electric field graphically by electric field lines — i.e., curves that represent the vector force exerted on a positive test charge.

- Electric field lines will originate on positive charges and terminate on negative charges.

- Electric field lines do not cross. (Why?) - The electric force at a given point in space is tangent to

the electric field line through that point. e) Properties of Field Lines - The strength of the electric field

is represented by the density of electric field lines

- The direction of the electric r

F

3 12. Electrostatics

field is tangent to the electric field lines

- The electric field lines from a point charge extend out radially.

- For a positive point charge, the field lines point outward and terminate at infinity

- For a negative charge, the field lines point inward and originate at infinity

f) Electric Field Lines for Two Point Charges - We can use the superposition principle to calculate the

electric field from two point charges.

- The field lines will originate from the positive charge and terminate on the negative charge.

g) Electric Field Lines from Identical Point Charges - For two positive charges, the field lines originate on the

positive charges and terminate at infinity.

- For two negative charges, the field lines terminate on the negative charges and originate at infinity.

h) Demo - visualization of electric field lines

- The charge of grass seeds is redistributed by induction.

- The Coulomb force makes the seeds align along the field lines.

12.3 Gauss’ law a) Electric Flux - The electric flux is defined to be

E = EA, Where E is the electric field and A is the area

- If surface area is not perpendicular to the electric field we have to slightly change our definition of the flux

E = EA cos Where f is the angle between the field and the unit vector that is perpendicular to the surface

- We can see that the relationship between the flux and the electric field and the area vector is just the dot product of two vectors

nAE

AE

E

E

ˆ

Where n̂ is a unit vector perpendicular to the surface

- The direction of a unit vector for an open surface is ambiguous

- For a closed surface, the unit vector is taken as being pointed outward

- Where flux lines enter the surface, the surface normal

and the electric field lines are anti-parallel - Where the flux lines exit the surface they are parallel - Is there a difference in the net flux through the cube

between the two situations?

- No! It is important to remember to properly take into

account the various dot products - The equation we have for flux is fine for simple situations

- the electric field is uniform and - the surface area is plane

- What happens when either one or the other or both is not true

- We proceed as we did in the transition from discrete charges to a continuous distribution of charges

- We break the surface area into small pieces and then calculate the flux through each piece and then sum them

- In the limit of infinitesimal areas this just becomes an

integral: AdEE

b) Electric Flux of a Point Charge

- We start from AdEE

,

- The electric field is given by 2

04

1

r

qE

- The problem has spherical symmetry, we therefore use a sphere as the Gaussian surface

- Since E is radial, its dot product with the differential area vector, which is also radial, is always one

- Also E is the same at every point on the surface of the sphere

- For these reasons, E can be pulled out from the integral and what

remains is dAEE

- The integral over the surface area of the sphere yields

- Pulling all this together then yields

E

E

rr

kqˆ)(

2xE

2 4 rA

4 12. Electrostatics

Example 4: A positive charge is contained inside a spherical shell. How does the differential electric flux, dФE, through the surface element dS change when the charge is moved from position 1 to position 2?

i) Increases ii) decreases iii) doesn’t change - The total flux of a charge is constant, with the density of

flux lines being higher the closer you are to the charge, Therefore as you move the charge closer to the surface element, the density of flux lines increases

- Multiplying this higher density by the same value for the area of dS gives us that the incremental flux also increases

Example 5: - A positive charge is contained inside a spherical shell. How

does the total flux, ФE, through the entire surface change when the charge is moved from position 1 to position 2?

i) ФE increases ii) ФE decreases iii) ФE doesn’t change - As we previously calculated, the total flux from a point

charge depends only upon the charge, so the total flux, ФE doesn’t change

c) Gauss’ Law - Gauss’ Law states that the net flux through any closed

surface equals the net (total) charge inside that surface divided by e0

- 0

netE

QAdE

- Note that the integral is over a closed surface - The result for a single charge can be extended to systems

consisting of more than one charge - One repeats the calculation for each of the charges

enclosed by the surface and then sum the individual

fluxes,

i

iE q0

1

- Gauss’ Law relates the flux through a closed surface to charge within that surface

Example 6: - A blue sphere A is contained within

a red spherical shell B. There is a charge QA on the blue sphere and charge QB on the red spherical shell.

- The electric field in the region between the spheres is completely

independent of QB the charge on the red spherical shell. Example 7: (Thin Infinite Sheet of Charge) - A given sheet has a charge density given

by s C/m2

- By symmetry, E is perpendicular to the sheet

- Use a surface that exploits this fact - A cylinder - A Gaussian pillbox

0

0

AAEEAAE

AAdE

rightcurvedleft

- But E and Acurved are perpendicular to each other so their dot product is zero and the middle term on the

left disappears 0

2

AAE ;

02

E

Example 8: (Infinite Line having a Charge Density l) - By Symmetry - E-field must be ^ to line of

charge and can only depend on distance from the line

- Therefore, choose the Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis

- Apply Gauss’ Law:

On the ends, 0 SdE

, since E// is zero

On the barrel, rhESdE 2

and hq

Equating these and rearranging yields r

E02

This is the same result as using the integral formulation Example 9: (Solid Uniformly Charged Sphere) - A charge Q is uniformly

distributed throughout the volume of an insulating sphere of radius R. Calculate average

charge density 3/R4

Q

3

Now select a Gaussian sphere of radius r within this larger sphere, the charge within this sphere is given by

3

33

3 3

4

3/4 R

rQr

R

QVQ enclencl

Electric Field is everywhere perpendicular to surface, i.e. parallel to surface normal Gauss’ Law then gives

30

3

3

0

2

0

4

4

R

rQE

R

rQrE

QAdE encl

Field increases linearly

0

2

20

44

1;

q

rr

qEA

E

EE

5 12. Electrostatics

within sphere Outside of sphere, electric field is given by that of a point charge of value Q

d) Charges on Conductors - Given a solid conductor, on which is placed an excess

charge then in the static limit - The excess charge will reside on the surface of the

conductor and - Everywhere the electric field due to this excess charge

will be perpendicular to the surface and - The electric field within the conductor will everywhere be

zero Example 10: - A solid conducting sphere is concentric

with a thin conducting shell, as shown - The inner sphere carries a charge Q1,

and the spherical shell carries a charge Q2, such that Q2 = - 3 Q1

(i) How is the charge distributed on the sphere?

- Remember that the electric field inside a conductor in a

static situation is zero.

- By Gauss’s Law, there can be no net

charge inside the conductor

- The charge, Q1, must reside on the

outside surface of the sphere

(ii) How is the charge distributed on the spherical shell?

- The electric field inside the conducting shell is zero.

- There can be no net charge inside the conductor

- Using Gauss’ Law it can be shown that the inner

surface of the shell must carry a net charge of –Q1

- The outer surface must carry the charge +Q1 + Q2, so

that the net charge on the shell equals Q2

- The charges are distributed uniformly over the inner

and outer surfaces of the shell, hence 2

2

1

4 R

Qinner

and 2

2

12

2

12

4

2

4 R

Q

R

QQouter

(iii) What is the electric field at r < R1? Between R1 and R2? And at r > R2?

- The electric field inside a conductor is zero.

- r < R1: This is inside the conducting sphere, therefore

0E

- Between R1 and R2 : R1 < r < R2

Charge enclosed within a Gaussian sphere = Q1

rr

QkE ˆ

21

- r > R2

Charge enclosed within a Gaussian sphere = Q1 + Q2

rr

Qkr

r

QQkr

r

QQkE ˆ

2ˆ

3ˆ

21

211

221

(iv) What happens when you connect the two spheres with a wire? (What are the charges?)

- After electrostatic equilibrium is

reached, there is no charge on the

inner sphere, and none on the inner

surface of the shell

- The charge Q1 + Q2 resides on the

outer surface

- Also, for r < R2, 0E

- and for r > R2, rr

QkE ˆ

221

Example 11: - An uncharged spherical conductor has a

weirdly shaped cavity carved out of it. Inside the cavity is a charge -q.

(i) How much charge is on the cavity wall? (a) Less than q (b) Exactly q (c) More than q By Gauss’ Law, since E = 0 inside the conductor, the

total charge on the inner wall must be q (and

therefore -q must be on the outside surface of the

conductor, since it has no net charge).

(ii) How is the charge distributed on the cavity wall? (a) Uniformly (b) More charge closer to –q (c) Less charge closer to –q The induced charge will distribute

itself non-uniformly to exactly cancel

everywhere in the conductor. The

surface charge density will be higher

near the -q charge.

(iii) How is the charge distributed on the outside of the sphere? (a) Uniformly (b) More charge near the cavity (c) Less charge near the cavity The charge will be uniformly distributed (because the

outer surface is symmetric). Outside the conductor,

the E field always points directly to

the center of the sphere, regardless

of the cavity or charge or its

location. Note: this is why your

radio, cell phone, etc. won’t work

inside a metal building!

12.4 Electrical potential 1) Electric Potential Energy - The electrostatic force is a conservative (=“path

independent”) force - It is possible to define an electrical potential energy

function with this force - Work done by a conservative force is equal to the

negative of the change in potential energy 2) Work and Potential Energy - There is a uniform field

between the two plates - As the positive charge

moves from A to B, work is done WAB=F d=q E d ΔPE =-W AB=-q E d

- only for a uniform field 3) Potential Difference

(=“Voltage Drop”) - The potential difference between points A and B is

defined as the change in the potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge

- ΔV = VB – VA = ΔPE /q - Potential difference is not the same as potential energy

6 12. Electrostatics - Another way to relate the energy and the potential

difference: ΔPE = q ΔV - Both electric potential energy and potential difference

are scalar quantities - Units of potential difference: V = J/C - A special case occurs when there is a uniform electric field

VB – VA= -Ed - Gives more information about units:

N/C = V/m 4) Energy and Charge Movements - A positive charge gains electrical potential energy when it

is moved in a direction opposite the electric field - If a charge is released in the electric field, it experiences a

force and accelerates, gaining kinetic energy - As it gains kinetic energy, it loses an equal amount of

electrical potential energy - A negative charge loses electrical

potential energy when it moves in the direction opposite the electric field

- When the electric field is directed downward, point B is at a lower potential than point A

- A positive test charge that moves from A to B loses electric potential energy

- It will gain the same amount of kinetic energy as it loses potential energy

5) When a positive charge is placed in an electric field - It moves in the direction of the field - It moves from a point of higher potential to a point of

lower potential - Its electrical potential energy decreases - Its kinetic energy increases 6) When a negative charge is placed in an electric field - It moves opposite to the direction of the field - It moves from a point of lower

potential to a point of higher potential

- Its electrical potential energy decreases

- Its kinetic energy increases 7) Electric Potential of a Point

Charge

- Note: if Q were a negative charge, V

would be negative 8) Electric Potential of Many Point Charges - Electric potential is a SCALAR not a vector. - Just calculate the potential due to

each individual point charge, and add together! (Make sure you get the SIGNS correct!)

- i i

i

r

qkV

9) Electric potential and electric potential energy

- U = Wapp = qV

- What is the potential energy of a dipole? - First bring charge +Q: no work involved, no potential

energy. - The charge +Q has created an electric potential

everywhere, V(r)= kQ/r - The work needed to bring the charge –Q to a distance a

from the charge +Q is Wapp=U = (-Q)V = (–Q)(+kQ/a) = -kQ

2/a

- The dipole has a negative potential energy equal to -kQ

2/a: we had to do negative work to build the dipole

(and the electric field did positive work). 10) Potential Energy of A System of Charges - 4 point charges (each +Q and equal

mass) are connected by strings, forming a square of side L

- If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart?

- Use conservation of energy: - Final kinetic energy of all four charges = initial potential

energy stored = energy required to assemble the system of charges

- No energy needed to bring in first charge: U1=0 - Energy needed to bring in 2nd charge:

- Energy needed to bring in 3rd charge =

- Energy needed to bring in 4th charge =

- Total potential energy is sum of all the individual terms =

- So, final kinetic energy of each charge = 244

2

L

kQ

Example 12:

- A proton moves from rest in an electric field of 8.0104

V/m along the +x axis for 50 cm. Find

a) the change in the electric potential,

b) the change in the electrical potential energy, and

c) the speed after it has moved 50 cm. (k = 8.99 X 109)

- a) V = -Ed = -(8.0104 V/m)(0.50 m) = -4.010

4 V

- b) PE = q V = (1.610-19

C)(-4.0 104 V) = -6.4 10

-15 J

- c) KEi + PEi = KEf + PEf, KEi = 0

KEf = PEi - PEf = -PE,

mpv2/2 = 6.410

-15 J

mp = 1.6710-27

kg

smkg

Jv /108.2

1067.1

)104.6(2 6

27

15

11) Electric Potential of a Point Charge - The point of zero electric potential is taken to be at an

infinite distance from the charge

7 12. Electrostatics - The potential created by a point charge q at any distance

r from the charge is r

qkV e ,

if r, V=0 and if r=0, V

12) Electric Potential of an electric Dipole

13) Electric Potential of Multiple Point Charges - Superposition principle applies - The total electric potential at some point P due to several

point charges is the algebraic sum of the electric potentials due to the individual charges

- The algebraic sum is used because potentials are scalar quantities

14) Electrical Potential Energy of Two Charges - V1 is the electric potential due to q1

at some point P1 - The work required to bring q2 from

infinity to P1 without acceleration is q2E1d = q2V1

- This work is equal to the potential energy of the two particle system

r

qqkVqPE e

2112

15) Notes About Electric Potential Energy of Two Charges - If the charges have the same sign, PE is positive

- Positive work must be done to force the two charges near one another

- The like charges would repel - If the charges have opposite signs, PE is negative

- The force would be attractive - Work must be done to hold back the unlike charges

from accelerating as they are brought close together

Example 12: - Finding the Electric Potential at Point P (apply V=keq/r).

V

mm

CCNmV

Vm

CCNmV

3

22

6229

2

46

2291

1060.3

)0.4()0.3(

)100.2()/1099.8(

,1012.10.4

100.5)/1099.8(

Superposition: Vp=V1+V2

Vp = 1.12104

V + (-3.60103 V) = 7.610

3 V

16) Problem Solving with Electric Potential (Point Charges) - Remember that potential is a scalar quantity - So no components to worry about - Use the superposition principle when you have multiple

charges - Take the algebraic sum - Keep track of sign - The potential is positive if the charge is positive and

negative if the charge is negative - Use the basic equation V = keq/r 17) Potentials and Charged Conductors - W = -DPE = -q(VB – VA) , no work is required to move a

charge between two points that are at the same electric

potential W=0 when VA=VB - All points on the surface of a charged conductor in

electrostatic equilibrium are at the same potential - Therefore, the electric potential is a constant everywhere

on the surface of a charged conductor in equilibrium 18) Conductors in Equilibrium (Overview) - The conductor has an excess of

positive charge - All of the charge resides at the surface - E = 0 inside the conductor - The electric field just outside the

conductor is perpendicular to the surface

- The potential is a constant everywhere on the surface of the conductor

- The potential everywhere inside the conductor is constant and equal to its value at the surface

19) The Electron Volt - The electron volt (eV) is defined as the energy that an

electron (or proton) gains when accelerated through a potential difference of 1 V

- Electrons in normal atoms have energies of 10’s of eV - Excited electrons have energies of 1000’s of eV - High energy gamma rays have energies of millions of eV

1 V=1 J/C 1 eV = 1.6 x 10-19

J 20) Electrical potential - The Coulomb force is a conservative force - A potential energy function can be defined for any

conservative force, including Coulomb force - The notions of potential and potential energy are

important for practical problem solving 21) Potential difference and electric potential - The electrostatic force is conservative

- As in mechanics, work is W = Fd cos

8 12. Electrostatics - Work done on the positive

charge by moving it from A

to B, W = Fd cos = qEd 22) Potential energy of

electrostatic field - The work done by a

conservative force equals the negative of the change in potential energy, DPE

PE = -W = -qEd - This equation - is valid only for the case of a uniform electric field - allows to introduce the concept of electric potential 23) Electric potential - The potential difference between points A and B, VB - VA,

is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by

the charge V = VB – VA = PE / q - Electric potential is a scalar quantity - Electric potential difference is a measure of electric

energy per unit charge 24) Potential is often referred to as “voltage” 25) Electric potential – units - Electric potential difference is the work done to move a

charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential

- In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V

- Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter)

- Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus,

- A positive charge gains electric potential energy when it is moved in a direction opposite the electric field

- A negative charge loses electrical potential energy when it moves in the direction opposite the electric field

26) Analogy between electric and gravitational fields - A positive charge gains electric potential energy when it is

moved in a direction opposite the electric field - A negative charge loses electrical potential energy when

it moves in the direction opposite the electric field - The same kinetic-potential energy theorem works here

- If a positive charge is released from A, it accelerates in

the direction of electric field, i.e. gains kinetic energy - If a negative charge is released from A, it accelerates

opposite the electric field Example 12: - motion of an electron - What is the speed of an electron accelerated from rest

across a potential difference of 100V? What is the speed of a proton accelerated under the same conditions? Given:

DV=100 V, me = 9.1110-31

kg

mp = 1.67 1́0-27

kg

|e| = 1.6010-19

C ve=? vp=?

- Observations: - given potential energy

difference, one can find the kinetic energy difference

- kinetic energy is related to speed 27) Electric potential and potential energy due to point

charges - Electric circuits: point of zero potential is defined by

grounding some point in the circuit - Electric potential due to a point charge at a point in

space: point of zero potential is taken at an infinite distance from the charge, Ve = Keq/r

- With this choice, a potential can be found as - Note: the potential depends only on charge of an object,

q, and a distance from this object to a point in space, r. 28) Superposition principle for potentials - If more than one point charge is present, their electric

potential can be found by applying superposition principle

- The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges.

- Remember that potentials are scalar quantities! 29) Potential energy of a system of point charges - Consider a system of two particles - If V1 is the electric potential due to charge q1 at a point P,

then work required to bring the charge q2 from infinity to P without acceleration is q2V1. If a distance between P and q1 is r, then by definition

- Potential energy is positive if charges are of the same sign and vice versa.

Example 12: potential energy of an ion - Three ions, Na+, Na+, and Cl-,

located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na+ ions?

30) Potentials and charged conductors - Recall that work is opposite of the change in potential

energy, - No work is required to move a charge between two

points that are at the same potential. That is, W=0 if VB=VA … but that’s not all!

- Recall:

a) All charge of the charged conductor is located on its surface

b) electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surface

- Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium

9 12. Electrostatics - Because the electric field in zero inside the conductor, no

work is required to move charges between any two

points, i.e. - If work is zero, any two points inside the conductor have

the same potential, i.e. potential is constant everywhere inside a conductor

- Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface!

- Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero!

31) The electron volt - A unit of energy commonly used

in atomic, nuclear and particle physics is electron volt (eV)

- The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V

- Relation to SI:

- 1 eV = 1.6010-19

C·V = 1.6010-19

J 32) Problem-solving strategy - Remember that potential is a scalar quantity, where

a) Superposition principle is an algebraic sum of potentials due to a system of charges

b) Signs are important - Just in mechanics, only changes in electric potential are

significant, hence, the point you choose for zero electric potential is arbitrary.

Example : ionization energy of the electron in a hydrogen atom

- In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x 10

-11 m. Find the ionization energy of the atom,

i.e. the energy required to remove the electron from the atom.

- Note that the Bohr model, the idea of electrons as tiny

balls orbiting the nucleus, is not a very good model of the

atom. A better picture is one in which the electron is

spread out around the nucleus in a cloud of varying

density; however, the Bohr model does give the right

answer for the ionization energy

- Given: r = 5.292 10-11

m, me = 9.1110-31

kg, mp =

1.6710-27

kg, |e| = 1.6010-19

C

Find: E=?

The ionization energy equals to the total energy of the

electron-proton system,

E = PE + KE with

The velocity of e can be found by analyzing the force on

the electron. This force is the Coulomb force; because the

electron travels in a circular orbit, the acceleration will be

the centripetal acceleration:

Thus, the total energy is

12.5 Equipotential surfaces (Ex) 1) Equipotentials and Electric Fields

Lines (Positive Charge) - The equipotentials for a point charge

are a family of spheres centered on the point charge

- The field lines are perpendicular to the electric potential at all points

It is convenient to represent by drawing equipotential lines

2) Equipotential Surfaces - An equipotential surface is a surface on which all points

are at the same potential - No work is required to move a charge at a constant speed

on an equipotential surface - The electric field at every point on an equipotential

surface is perpendicular to the surface - defined as a surface in space on which the potential is the

same for every point (surfaces of constant voltage) - The electric field at every point of an equipotential

surface is perpendicular to the surface

3) Applications: Electrostatic Precipitator - It is used to remove particulate

matter from combustion gases - Reduces air pollution - Can eliminate approximately 90%

by mass of the ash and dust from smoke

- High voltage (4-100 kV) is maintained between the coil wire and the grounded wall

- The electric field at the wire causes discharges, i.e., ions (charged oxygen atoms) are formed

- The negative ions and electrons move to the positively biased wall

- On their way the ions and electrons ionize dirt particles due to collisions

- Most of the dirt particles become negatively charged and are attracted to the wall as well – cleaning effect

4) Applications: Electrostatic Air Cleaner - Used in homes to relieve the discomfort of allergy

sufferers - It uses many of the same principles as the electrostatic

precipitator 5) Application: Xerographic Copiers - The process of xerography is used for making photocopies - Uses photoconductive materials - A photoconductive material is a poor conductor of

electricity in the dark but becomes a good electric conductor when exposed to light

10 12. Electrostatics - The Xerographic Process:

6) Application: Laser Printer - The steps for producing a document on a laser printer is

similar to the steps in the xerographic process - Steps a, c, and d are the same - The major difference is the way the image forms of the

selenium-coated drum - A rotating mirror inside the printer causes the beam of

the laser to sweep across the selenium-coated drum - The electrical signals form the desired letter in positive

charges on the selenium-coated drum - Toner is applied and the process continues as in the

xerographic process 7) Point of Equilibrium - Clearly, half way between two equal charges is a point of

equilibrium, P, as shown on the left in the diagram. (This means there is zero net force on any charge placed at P.) At no other point in space, even points equidistant between the two charges, will equilibrium occur.

- Depicted on the right are two positive point charges, one

with twice the charge of the other, separated by a distance d. In this case, P must be closer to q than 2 q since in order for their forces to be the same, we must be closer to the smaller charge. Since Coulomb’s formula is nonlinear, we can’t assume that P is twice as close to the smaller charge. We’ll call this distance x and calculate it in terms of d.

- Since P is the equilibrium point, no matter what charge is placed at P, there should be zero electric on it. Thus an arbitrary “test charge” q0 (any size any sign) at P will feel a force due to q and an equal force due to 2q. We compute each of these forces via Coulomb’s law:

The K’s, q’s, and q0’s cancel, the latter showing that the location of P is independent of the charge placed there. Cross multiplying we obtain:

(d - x)2 = 2 x

2

d

2 - 2 x d + x

2 = 2 x

2

x 2 +

2 x d - d

2 = 0.

From x 2 +

2 x d - d

2 = 0, the quadratic formula yields:

Since x is a distance, we choose the positive root:

x = d ( 2 - 1 ) 0.41 d. Note that x < 0.5 d, as predicted. Note that if the two charges had been the same, we would have started with

(d - x)2 = x

2 d

2 - 2 x d + x

2 = x

2

d 2 -

2 x d = 0 d (d - 2 x ) = 0 x = d / 2, as

predicted. This serves as a check on our reasoning.

8) Equilibrium with Several Charges 9) Several equal point charges are to be arranged in a

plane so that another point charge with non-negligible mass can be suspended above the plane. How might this be done?

- Arrange the charges in a circle, spaced evenly, and fix

them in place. Place another charge of the same sign above the center of the circle. If placed at the right distance above the plane, the charge could hover. This arrangement works because of symmetry. The electric force vectors on the hovering charge are shown. Each vector is the same magnitude and they lie in a cone. Each vector has a vertical component and a component in the plane.

- The planar components cancel out, but the vertical components add to negate the weight vector.

- If the charges in the plane are arranged in a circle with a large radius, the electric force vectors would be more horizontal, thereby working together less and canceling each other more. The hovering charge would lower. Since its weight doesn’t change, it must be closer to the plane in order to increase the forces to compensate for their partial cancellation. If the charges in the plane were arranged in a small circle, the vectors would be more vertical, thereby working together more and canceling each other less. The hovering charge would rise and the vectors would decrease in magnitude. To maximize the height of the hovering charge, all the charges in the plane should be brought to a single point.

11 12. Electrostatics Summary

Electrostatics

Coulomb’s Law

F = Q1Q2/(40r2)

Electric Potential, V

V = -r Edx

E = - dV/dx

E = - dV/dr

U = qV

Electric Field

E = F/q

Motion of point charge in

uniform electric field E.

Gauss’s Law

= Q/0

= EA

Uniform charged palate:

E = /20

Point charge: E = Q/(40r2)

Charged sphere:

For r > R, E = 0

For r R, E = Q/(40r2)