Physics STPM - Chung Hua Middle School STPM Community

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 /
2
4πε0r
(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
2
(g) use the formula V = Q / 4πε0r
(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
-19
charge is e=1.6×10 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.
qq
F  1 2 2 (Newtons)
4o r
where:
q = charge, measured in
Coulombs (C).
k = Coulombs constant
9
2 2
= 8.99 x 10 N.m /C .
qq
- Or F  1 2 2 , where
4o r
k=
1
40
 8.99  109
Nm 2
C2
,
ε0 = permittivity constant = 8.85  1012
C2
Nm 2
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
- gravitation is always attractive (no
QQ
Fe  k 1 2 2
negative mass)
r
electrical force can be both
mm
Fg  G 1 2 2
attractive or repulsive
r
- Electric force is dominant in the
Nm2
atomic world
k  9  109 2
Gravitational forces dominates on
C
the macroscopic scale: people,
Nm2
planets, galaxies
G  6.7  1011
kg 2
- Electric forces are more stronger !
Example 1:
What is the electrostatic force between two positive
charges? [q1 = 4 μC, q2 = 8 μC, r = 10 cm]
q1
F
2
q2
r
9
-6
F
-6
2
F = kqq/r = (9 x 10 ) (4 x 10 ) (8 x 10 ) / (0.1) = 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.
2 12. Electrostatics
Example 2:
Forces F1 and F2 act independently on test charge (q0).
[q1 = 2 μC, q2 = 5 μC, q0 = 4 cm]
2
F1 = kqq/r = 7.2 N (to right)
2
F2 = kqq/r = 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.
2
9
F=kq1q2 / r (k=9 x 10 , q=charge of particle, r=distance
between particles)
Force due to q1:
9
-9
-9
2
F1=(9 x 10 )(5.46 x 10 )(4.95 x 10 ) / (0.3)
-6
F1 = 2.7027 x 10 N to the left since repulsive force
Force due to q3:
9
-9
-9
2
F2=(9 x 10 )(4.95 x 10 )(2.97 x 10 ) / (0.1)
-5
F2 = 1.323 x 10 N downwards since attractive force
To find magnitude, use pythagorean theorem
2
-6 2
-5 2
F = (2.7027 x 10 ) + (1.323 x 10 )
-5
F = 1.35 x 10 N
a)
Find the direction of the electric force on the particle at
the origin.
To find direction, use
trigonometry
-5
tan  = 1.323 x 10 /
-6
2.7027 x 10
 = 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

 F
force that acts on the test charge; then E 
q
- 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 F  qE (x )
- 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.

qq
- Force on q0: F  k 20 rˆ
r

 
F
q
 k 2 rˆ
- Electric field at x: E ( x ) 
q0
r
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
F
is represented by the density of
electric field lines
r
- The direction of the electric
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
kq
E(x )  2 rˆ
inward and originate at infinity
r
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
 
E  E  A

 E  E  Anˆ
Where n̂ is a unit vector perpendicular to the surface
- The direction of a unit vector for an open surface is
ambiguous
E
E
- 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:  E  E  dA

b) Electric Flux of a Point Charge
 
- We start from  E  E  dA ,

- The electric field is given by E 
-
1
q
40 r 2
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  E  E dA
- The integral over the surface area
of the sphere yields
A  4 r 2
- Pulling all this together then yields
4 12. Electrostatics
independent of QB the charge on the red spherical shell.
 E  EA;  E 
E 
1
q
4 r
40 r 2
2
q
0
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?
Example 7: (Thin Infinite Sheet of Charge)
- A given sheet has a charge density given
2
by s C/m
- By symmetry, E is perpendicular to the
sheet
- Use a surface that exploits this fact
- A cylinder
- A Gaussian pillbox


A
 E  dA   0
E Aleft  EAcurved  E Aright 
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?
A
0
- But E and Acurved are
perpendicular to each other
so their dot product is zero and the middle term on the
A

left disappears 2 E A 
; E
0
2 0
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, E  dS  0 , since E// is zero
 
On the barrel, E  dS  2 rhE and q  h

Equating these and rearranging yields E 
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
  Q
-  E  E  dA  net

0
- 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
1
qi
fluxes,  E 
0 
i
- 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

20 r
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
Q
charge density  
4 R 3 / 3
Now select a Gaussian sphere of radius r within this larger
sphere, the charge within this sphere is given by
3

 4 3
Q
  r   Q r
Qencl   Vencl  
 4 R 3 / 3   3

R3


Electric Field is everywhere perpendicular to surface, i.e.
parallel to surface normal
Gauss’ Law then gives
  Q
E  dA  encl

0
E 4 r 2 
E
Q
Q r3
 0 R3
r
40 R 3
Field increases linearly
5 12. Electrostatics
d)
-
within sphere
Outside of sphere, electric field is given by that of a point
charge of value Q
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
Q1
and outer surfaces of the shell, hence  inner  
4R2 2
Q Q
2Q1
and  outer  2 21 
4R2
4R2 2
(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

E 0
- Between R1 and R2 : R1 < r < R2
Charge enclosed within a Gaussian sphere = Q1

Q
E  k 21 rˆ
r
- r > R2
Charge enclosed within a Gaussian sphere = Q1 + Q2

Q Q
Q  3Q
2Q
E  k 1 2 2 rˆ  k 1 2 1 rˆ  k 21 rˆ
r
r
r
(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, E  0

2Q
- and for r > R2, E  k 21 rˆ
r
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?
(ii)
(iii)
(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).
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.
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!)
q
- V k i
ri
i

9) Electric potential and electric potential energy
- U = Wapp = qV
- 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
2
Wapp=U = (-Q)V = (–Q)(+kQ/a) = -kQ /a
- The dipole has a negative potential energy equal to 2
kQ /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 =

2
- So, final kinetic energy of each charge = kQ 4  2
4L

Example 12:
4
- A proton moves from rest in an electric field of 8.010
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)
4
4
- a) V = -Ed = -(8.010 V/m)(0.50 m) = -4.010 V
-19
4
-15
- b) PE = q V = (1.610 C)(-4.0 10 V) = -6.4 10 J
- c) KEi + PEi = KEf + PEf, KEi = 0
KEf = PEi - PEf = -PE,
2
-15
mpv /2 = 6.410 J
-27
mp = 1.6710 kg
v
2(6.4  1015 J )
1.67  1027 kg
 2.8  106 m / s
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
q
r from the charge is V  ke ,
r
if r, V=0 and if r=0, V 
V1  (8.99  109 Nm2 / C 2 )
V2  (8.99  109 Nm2 / C 2 )
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
qq
PE  q2V1  ke 1 2
r
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).
5.0  106 C
 1.12  104V ,
4.0m
( 2.0  106 C )
(3.0m)  (4.0m)
2
2
 3.60  103V
Superposition: Vp=V1+V2
4
3
3
Vp = 1.1210 V + (-3.6010 V) = 7.610 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
-19
1 V=1 J/C  1 eV = 1.6 x 10 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:
-31
DV=100 V, me = 9.1110 kg
-27
mp = 1.67´10 kg
-19
|e| = 1.6010 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
1V
- Relation to SI:
-19
-19
- 1 eV = 1.6010 C·V = 1.6010 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
-11
= 5.29 x 10 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 = 9.1110-31 kg, mp =
1.6710-27 kg, |e| = 1.6010-19 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:
2
2
(d - x) = 2 x
2
2
2
 d -2xd+x =2x
2
2
 x + 2 x d - d = 0.
2
2
From x + 2 x d - d = 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
2
2
2
2
2
(d - x) = x  d - 2 x d + x = x
2
 d - 2 x d = 0  d (d - 2 x ) = 0  x = d / 2, as
predicted. This serves as a check on our reasoning.
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:
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
Electric Field
Gauss’s Law
F=
Q1Q2/(40r2)
E = F/q
 = Q/0
Motion of point
charge in
uniform electric
field E.
 = EA
Electric
Potential, V
Uniform charged
palate:
V = -r Edx
E = /20
E = - dV/dx
Point charge: E =
Q/(40r2)
E = - dV/dr
U = qV
Charged sphere:
For r > R, E = 0
For r  R, E =
Q/(40r2)