Transcript The Electric Field

Welcome…
…to Electrostatics
Outline
1. Coulomb’s Law
2. The Electric Field
- Examples
3. Gauss Law
- Examples
4. Conductors in Electric Field
Coulomb’s Law
Coulomb’s law quantifies the magnitude of the electrostatic
force.
Coulomb’s law gives the force (in Newtons) between charges q1
and q2, where r12 is the distance in meters between the
charges, and k=9x109 N·m2/C2.
q1q 2
F k 2
12
r12
Coulomb’s Law
Force is a vector quantity. The equation on the previous slide
gives the magnitude of the force.
If the charges are opposite in sign, the force is attractive;
if the charges are the same in sign, the force is repulsive.
Also, the constant k is equal to 1/40,
where 0=8.85x10-12 C2/N·m2.
F12 
1
q1q2
40 r
2
12
Coulomb’s Law
• If q1 and q2 are located at
points having position vectors
r1 and r2 then the Force F12 is
given by;
F12 
1
q1q2
40 r
R12  r2  r1
2
12
a R12
The equation is valid for point charges. If the charged objects
are spherical and the charge is uniformly distributed, r12 is the
distance between the centers of the spheres.
r12
-
+
If more than one charge is involved, the net force is the vector
sum of all forces (superposition). For objects with complex
shapes, you must add up all the forces acting on each separate
charge (turns into calculus!).
+
+
+
-
-
-
Solving Problems Involving Coulomb’s Law and
Vectors
Example: Calculate the net electrostatic force on charge Q3
due to the charges Q1 and Q2.
y
30 cm
Q3=+65C
=30º
Q1=-86C
Q2=+50C
52 cm
x
Step 0: Think!
This is a Coulomb’s Law problem (all we have to work with, so
far).
We only want the forces on Q3.
Forces are additive, so we can calculate F32 and F31 and add
the two.
If we do our vector addition using components, we must resolve
our forces into their x- and y-components.
Step 1: Diagram
y
Draw and label relevant
quantities.
Draw axes, showing
origin and directions.
30 cm
F32
Draw a representative
sketch.
Q3=+65C
F31
=30º
Q1=-86C
Q2=+50C
52 cm
Draw and label forces (only those on Q3).
Draw components of forces which are not along axes.
x
Step 2: Starting Equation
y
F32
30 cm
Q3=+65C
F31
=30º
Q1=-86C
Q2=+50C
52 cm
q1q 2
F k 2
12
r12
“Do I have to put in the absolute value signs?”
x
Step 3: Replace Generic Quantities by Specifics
y
repulsive
Q3 Q 2
F
k 2
32, y
r32
Q3=+65C
r32=30 cm
Q3Q 2
F k 2 ,
32
r32
F32
F31
=30º
Q1=-86C
Q2=+50C
F
 0 (from diagram)
32, x
F32,y = 330 N and F32,x = 0 N.
52 cm
x
Step 3 (continued)
attractive
Q3Q1
F
  k 2 cos 
31, x
r31
(+ sign comes from
diagram)
Q3Q1
F
 k 2 sin 
31, y
r31
y
F32
Q3=+65C
r32=30 cm
Q3Q1
F k 2 ,
31
r31
F31
=30º
Q1=-86C
Q2=+50C
52 cm
(- sign comes from diagram)
You would get F31,x = +120 N and F31,y = -70 N.
x
Step 3: Complete the Math
y
Q3=+65C
30 cm
The net force is the
vector sum of all the
forces on Q3.
F32
F3
F31
=30º
Q1=-86C
Q2=+50C
52 cm
F3x = F31,x + F32,x = 120 N + 0 N = 120 N
F3y = F31,y + F32,y = -70 N + 330 N = 260 N
You know how to calculate the magnitude F3 and the angle
between F3 and the x-axis.
x
Many Point Charges
• If we have N point charges Q1, Q2… QN, located at r1, r1…
rN the vector sum of the forces exerted on Q by each of the
charges is given by
F
Q
40
N

k 1
Qk (r  rk )
r  rk
3
Coulomb’s Law:
The Big Picture
Coulomb's Law quantifies the interaction between charged
particles.
1 q1q 2
F =
,
2
12 4πε 0 r12
r12
+
-
Q1
Q2
Coulomb’s Law was discovered through decades of experiment.
By itself, it is just “useful." Is it part of something bigger?
The Electric Field
Coulomb's Law (demonstrated in 1785)
shows that charged particles exert forces
on each other over great distances.
How does a charged particle "know"
another one is “there?”
The Electric Field
Faraday, beginning in the 1830's, was the leader in developing
the idea of the electric field. Here's the idea:
F12
 A charged particle emanates a "field"
into all space.
 Another charged particle senses the field,
and “knows” that the first one is there.
+
+
F21
like
charges
repel
F13
F31
unlike
charges
attract
The Electric Field
We define the electric field by the
force it exerts on a test charge
q0:
F0
E=
q0
By convention the direction of the electric
field is the direction of the force exerted
on a POSITIVE test charge. The absence
of absolute value signs around q0 means
you must include the sign of q0 in your
work.
The Electric Field
If the test charge is "too big" it perturbs the electric field, so the
“correct” definition is
F0
E = lim
q 0 0 q
0
You won’t be required to use
this version of the equation.
Any time you know the electric field, you can use this equation
to calculate the force on a charged particle in that electric field.
F = qE
The units of electric field are
Newtons/Coulomb.
 F0  N
 E  =   =
q0  C
Later you will learn that the units of electric field can also be
expressed as volts/meter:
N V
E = =
C m
The electric field exists independent of whether there is a
charged particle around to “feel” it.
Remember: the electric field direction is the
direction a + charge would feel a force.
+
A + charge would be repelled by another + charge.
Therefore the direction of the electric field is away from positive
(and towards negative).
The Electric Field
Due to a Point Charge
Coulomb's law says
q1q 2
F =k 2 ,
12
r12
... which tells us the electric field due to a point charge q is
E q =k
q
r
2
, away from +
…or just…
This is your third starting equation.
q
E=k 2
r
We define r̂ as a unit vector from the source point to the field
point:
source point
r̂ +
field point
The equation for the electric field of a point charge then
becomes:
q
E=k 2 rˆ
r
You may start with either equation
for the electric field (this one or the
one on the previous slide). But
don’t use this one unless you
REALLY know what you are doing!
Many Point Charges
• If we have N point charges Q1, Q2… QN, located at r1, r1…
rN the vector sum of the Electric Field E by each of the
charges is given by
E
1
40
N

k 1
Qk (r  rk )
r  rk
3
Motion of a Charged Particle
in a Uniform Electric Field
A charged particle in an electric field experiences a force, and if
it is free to move, an acceleration.
If the only force is due to the
electric field, then
 F  ma  qE.
- - - - - - - - - - - - -
-
F
E
+ + + + + + + + + + + + +
If E is constant, then a is constant, and you can use the
equations of kinematics.
Example: an electron moving with velocity v0 in the positive x
direction enters a region of uniform electric field that makes a
right angle with the electron’s initial velocity. Express the
position and velocity of the electron as a function of time.
y
- - - - - - - - - - - - -
x
-
v0
E
+ + + + + + + + + + + + +
The Electric Field
Due to a Collection of Point Charges
The electric field due to a small "chunk" q of charge is
1 q
E =
r
2
4πε 0 r
unit vector from q
to wherever you
want to calculate E
The electric field due to collection of "chunks" of charge is
1
E =  E i =
4πε 0
i
q i
i r 2 r i
i
As qdq0, the sum becomes an integral.
Charge Distributions
• Continious charge distributions are given by;
• Do not confuse ρ with radial distance
Line Charge
If charge is distributed along a
straight line segment parallel to
the z-axis, the amount of charge
dq on a segment of length dz is
ρ dz.
The Charge elemen is given by;
The total charge is given by;
Line Charge
The electric field at point P due
to the charge dq is
1 dq
1 dx
dE =
r' =
r'
2
2
4πε 0 r'
4πε 0 r'
The field point is denoted by (x,y,z) while
the source point is denoted by (x’,y’,z’)
I’m assuming positively charged objects
in these “distribution of charges” slides.
Line Charge
Definitions
Line Charge
• Using given definitions we have;
• For a finite line charge;
Line Charge
• For an infinite line charge α1 and α2 are equal to
π/2 and - π/2 respectively, the z component is
canceled and the equation becomes
Surface Charge
If charge is distributed over a two-dimensional surface, the
amount of charge dq on an infinitesimal piece of the surface is
ρs dS, where ρs is the surface density of charge (amount of
charge per unit area).
y
ρs
charge dQ = ρs dS
x
area = dS
Surface Charge
y
P
dE
r’
ar
The electric field at P due to the charge dq is
x
Surface Charge
P
dE
r’
ar
x
The net electric field at P due to the entire surface of charge is
E
1
40

s
 S dS
r'
2
aR
Surface Charge
• Definitions
• And the E field becomes;
Surface Charge
• After evaluating the previous integration;
• Or a more general formula is
an is a unit vector
normal to the
sheet
• Note that the Electric Field is independent of the
distance between the sheet and the point of
observation P
Volume Charge
The net electric field at P due to a three-dimensional
distribution of charge is…
z
E
P
r’
r'
x
y
1
(x, y, z) dV
E=
r'
.
2

4πε 0 V
r'
Volume Charge
• For a sphere with radius a
centered at the origin
• The E field due to elementary
volume charge is given by;
Volume Charge
• Due to the symmetry of the
charge distribution, the
contributions to Ex or Ey add
up to zero
Volume Charge
• Definitions
• And the resulting expression for
E field is;
Summarizing:
Charge distributed along a line:
1
λ dx
E=
r' 2 .

4πε 0
r'
Charge distributed over a surface:
1
 dS
E=
r' 2 .

4πε 0 S r'
Charge distributed inside a volume:
1
 dV
E=
r' 2 .

4πε 0 V r'
If the charge distribution is uniform, then , , and  can be taken outside
the integrals.
The Electric Field
Due to a Continuous Charge Distribution
(worked examples)
Example: A rod of length L has a uniform charge per unit length
 and a total charge Q. Calculate the electric field at a point P
along the axis of the rod at a distance d from one end.
y
P
x
d
L
Let’s put the origin at P. The linear charge density and Q are
related by
Q
=
and Q = L
L
Let’s assume Q is positive.
y
dE
x
P
d
dQ =  dx
dx
x
L
The electric field points away from the rod. By symmetry, the
electric field on the axis of the rod has no y-component. dE
from the charge on an infinitesimal length dx of rod is
dq
 dx
dE = k 2  k 2
x
x
Note: dE is in the –x direction. dE is the magnitude of dE. I’ve
used the fact that Q>0 (so dq=0) to eliminate the absolute
value signs in the starting equation.
y
dE
x
P
d
E =
d+L
d
dQ =  dx
dx
x
L
dE x = -k 
d+L
d
 dx ˆ
i = -k 
2
d
x
d+L
dL
dx ˆ
 1 ˆ
i = -k    i
2
x
 x d
 d  d  L  ˆ
1
1 ˆ
L ˆ
kQ ˆ

E = -k  
  i = -k 
i=
-k
i=
i

d d  L
d d  L
 dL d
 d d  L 
Example: A ring of radius a has a uniform charge per unit
length and a total positive charge Q. Calculate the electric field
at a point P along the axis of the ring at a distance x0 from its
center.
dQ
r
a

x0
P

dE
x
By symmetry, the y- and zcomponents of E are zero,
and all points on the ring
are a distance r from point
P.
dQ
dQ
dE=k 2
r
r
a

x0
P

x
No absolute value
signs because Q is
positive.
dQ
dE x =k 2 cos 
r
dE
r = x a
2
0
2
x0
cos  
r
x0
 dQ  x 0
E x   dE x    k 2   k 3
r  r
r
ring
ring 
For a given x0, r is a constant
for points on the ring.
x0
kx 0Q
ring dQ  k r3 Q  x 2  a 2 3/ 2


Or, in general, on the ring axis E x,ring 
0
kxQ
2
x
 a

2 3/ 2
.
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
dQ
r
P
R
x
x0
The disc is made of
concentric rings. The
area of a ring at a
radius r is 2rdr, and
the charge on each ring
is (2rdr).
We can use the equation on the previous slide for the electric
field due to a ring, replace a by r, and integrate from r=0 to
r=R.
dE ring 
kx 0 2rdr
x
2
0
r

2 3/ 2
.
Caution! I’ve switched
the “meaning” of r!
dQ
r
P
x
x0
R
Ex 
 dE
disc
x


disc
kx 0 2rdr
x
 x2  r 
0
E x  kx 0  
 1/ 2

2
0
r
2 1/ 2

2 3/ 2
R
 kx 0  
R
0
x
2r dr
2
0
r

2 3/ 2



x
x
0

  2k  0 
 x 0  x 2  R 2 1/ 2 

0
0


Example: Calculate the electric field at a distance x0 from an
infinite plane sheet with a uniform charge density .
Treat the infinite sheet as disc of infinite radius.
1
Let R and use k 
to get
4 0
Esheet 

20
.
Interesting...does not depend on distance from the sheet.
I’ve been Really Nice and put this on your starting equation sheet. You don’t have to
derive it for your homework!
Electric Field Lines
Electric field lines help us visualize the electric field and predict
how charged particles would respond the field.
-
+
Example: electric field lines for isolated +2e and -e charges.
Here’s how electric field lines are related to the field:
 The electric field vector E is tangent to the field lines.
 The number of lines per unit area through a surface
perpendicular to the lines is proportional to the electric field
strength in that region
 The field lines begin on positive charges and end on
negative charges.
 The number of lines leaving a positive charge or
approaching a negative charge is proportional to the
magnitude of the charge.
 No two field lines can cross.
Example: draw the electric field lines for charges +2e and -1e,
separated by a fixed distance. Easier to use this link!
Gauss’ Law
Electric Flux
We have used electric field lines to visualize electric fields and
indicate their strength.
We are now going to count the
number of electric field lines passing
through a surface, and use this
count to determine the electric field.
E
Electric Flux
The electric flux passing through a surface is the number of
electric field lines that pass through it.
Because electric field lines are drawn
arbitrarily, we quantify electric flux
like this: E=EA, except that…
A
If the surface is tilted, fewer lines cut
the surface.
E
E

Electric Flux
We define A to be a vector having a
magnitude equal to the area of the
surface, in a direction normal to the
surface.
A

E

The “amount of surface” perpendicular
to the electric field is A cos .
Because A is perpendicular to the surface, the amount of A
parallel to the electric field is Acos .
A = A cos  so E = EA = EA cos .
Remember the dot product?
 E  E A
Electric Flux
If the electric field is not uniform, or the surface is not flat…
divide the surface into
infinitesimal surface
elements and add the
flux through each…
dA
A
E
 E   E  dA
Electric Flux
If the surface is closed (completely encloses a volume)…
…we count* lines going
out as positive and lines
going in as negative…
E
dA
 E   E  dA
a surface integral, therefore a
double integral 
Electric Flux
Use the simplest (easiest!) one that works.
 E  EA
Flat surface, E  A, E constant over surface. Easy!
 E  EA cos 
Flat surface, E not  A, E constant over surface.
E  E  A
Flat surface, E not  A, E constant over surface.
 E   E  dA
Surface not flat, E not uniform. Avoid, if possible.
E 
 E  dA
Closed surface. Most general. Most complex.
If the surface is closed, you may be able to “break it up” into
simple segments and still use E=E·A for each segment.
Gauss’ Law
Gauss's law states that the total electric flux  E through any
closed surface is equal to the total charge enclosed by that
surface.
Gauss’ Law
Gauss’ Law
Differential form of Gauss’ Law
the volume charge density is the same as the
divergence of the electric flux density.
Gauss’ Law
We will find that Gauss law
gives a simple way to calculate
electric fields for charge
distributions that exhibit a high
degree of symmetry…
Point Charge Example
Example: use Gauss’ Law to calculate the electric field from an
isolated point positive charge q.
To apply Gauss’ Law, we construct a “Gaussian Surface”
enclosing the charge.
The Gaussian surface should mimic the symmetry of the charge
distribution.
For this example, choose for our Gaussian surface a sphere of
radius r, with the point charge at the center.
Point Charge Example
Starting with Gauss’s law, calculate the electric
field due to an isolated point charge q.
r
q
E
dA
We choose a Gaussian surface that is a
sphere of radius r centered on the point
charge.
Since the charge is positive the field is radial
outward and everywhere perpendicular to the
Gaussian surface.
Gauss’s law then gives:
Symmetry tells us that the field is
constant on the Gaussian surface.
Strategy for Solving Gauss’ Law Problems
 Select a Gaussian surface with symmetry that matches the
charge distribution.
 Draw the Gaussian surface so that the electric field is either
constant or zero at all points on the Gaussian surface.
 Use symmetry to determine the direction of E on the Gaussian
surface.
 Evaluate the surface integral (electric flux).
 Determine the charge inside the Gaussian surface.
 Solve for E.
Example: use Gauss’ Law to calculate the electric field due to a
long line of charge, with linear charge density .
Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
These are easy using Gauss’ Law (remember what a pain they
were in the previous chapter). Study these examples and others
in your text!
E line


.
20 r
E sheet


.
20
Worked Example 1
Compute the electric flux through a cylinder with an axis parallel to the electric
field direction.
E
A
The flux through the curved surface is zero since E is
perpendicular to dA there. For the ends, the surfaces are
perpendicular to E, and E and A are parallel. Thus the flux
through the left end (into the cylinder) is –EA, while the flux
through right end (out of the cylinder) is +EA. Hence the
net flux through the cylinder is zero.
Gauss’s Law
Gauss’s Law relates the electric flux through a closed surface with
the charge Qin inside that surface.
E 
Qin
 E  dA  
0
This is a useful tool for simply determining the electric
field, but only for certain situations where the charge
distribution is either rather simple or possesses a high
degree of symmetry.
Worked Example 3
An insulating sphere of radius a has a uniform charge density ρ and a total
positive charge Q. Calculate the electric field outside the sphere.
Since the charge distribution is spherically
symmetric we select a spherical Gaussian
surface of radius r > a centered on the
charged sphere. Since the charged sphere
E
r
has a positive charge, the field will be
a
dA
directed radially outward. On the Gaussian
sphere E is always parallel to dA, and is
Q
constant.
 Left side:

Right side:
Qin
0

Q
0

 E  dA 
E  4 r
2
 
Q
0
 E dA  E

1

dA  E 4 r 2
Q
Q
or E 
 ke 2
2
4 0 r
r

Worked Example 3 cont’d
r
Find the electric field at a point inside the sphere.
Now we select a spherical Gaussian surface
with radius r < a. Again the symmetry of the
charge distribution allows us to simply evaluate
the left side of Gauss’s law just as before.
a

Q
Left side:
 E  dA 

2
E
dA

E
dA

E
4

r


The charge inside the Gaussian sphere is no longer Q. If we
call the Gaussian sphere volume V’ then

4 3

Right side: Qin   V    r 
3
Q
4

r
3
E 4 r 2  in 
0
3 0



4  r 3

Q
1 Q
Q
E

r but  
so E 
r  ke 3 r
3
2
4
3 0
4 0 a
a
3 0 4 r
 a3
3



Worked Example 3 cont’d
Q
We found for r  a , E  ke 2
r
ke Q
and for r  a , E  3 r
a
a
Q
E
Let’s plot this:
a
r
Conductors in Electrostatic
Equilibrium
By electrostatic equilibrium we mean a situation where
there is no net motion of charge within the conductor
The electric field is zero everywhere inside
the conductor
Any net charge resides on the conductor’s
surface
The electric field just outside a charged
conductor is perpendicular to the conductor’s
surface
Conductors in Electrostatic
Equilibrium
The electric field is zero everywhere inside the conductor
Why is this so?
If there was a field in the conductor the charges
would accelerate under the action of the field.
++++++++++++
---------------------
The charges in the conductor
move creating an internal
electric field that cancels the
applied field on the inside of
the conductor
Ein
E
E
The charge on the right is twice the magnitude of the
charge on the left (and opposite in sign), so there are
twice as many field lines, and they point towards the
charge rather than away from it.
Combinations of charges. Note that, while the lines are less
dense where the field is weaker, the field is not necessarily
zero where there are no lines. In fact, there is only one point
within the figures below where the field is zero – can you
find it?
The electric field is always perpendicular to the
surface of a conductor – if it weren’t, the
charges would move along the surface.
The electric field is stronger where the surface is
more sharply curved.
Summary
Two methods for calculating electric field
Coulomb’s Law
Gauss’s Law
Gauss’s Law: Easy, elegant method for
symmetric charge distributions
Coulomb’s Law: Other cases
Gauss’s Law and Coulomb’s Law are
equivalent for electric fields produced by
static charges
Worked Example 4
Any net charge on an isolated conductor must reside on its surface and
the electric field just outside a charged conductor is perpendicular to its
surface (and has magnitude σ/ε0). Use Gauss’s law to show this.
For an arbitrarily shaped conductor we
can draw a Gaussian surface inside the
conductor. Since we have shown that the
electric field inside an isolated conductor
is zero, the field at every point on the
Gaussian surface must be zero.
Qin
 E  dA  
0
From Gauss’s law we then conclude that the
net charge inside the Gaussian surface is
zero.
Since the surface can be made
arbitrarily close to the surface of the
conductor, any net charge must reside on the
conductor’s surface.
Worked Example 4 cont’d
We can also use Gauss’s law to determine the electric field just outside
the surface of a charged conductor. Assume the surface charge density
is σ.
Since the field inside the conductor is
zero there is no flux through the face
of the cylinder inside the conductor. If
E had a component along the surface
of the conductor then the free charges
would move under the action of the
field creating surface currents. Thus E
is perpendicular to the conductor’s
surface, and the flux through the
cylindrical surface must be zero.
Consequently the net flux through the
cylinder is EA and Gauss’s law gives:
A

 E  EA 

or E 
0
0
0
Qin
Worked Example 5
A conducting spherical shell of inner radius a and outer radius b with a
net charge -Q is centered on point charge +2Q. Use Gauss’s law to
find the electric field everywhere, and to determine the charge
distribution on the spherical shell.
First find the field for 0 < r < a
This is the same as Ex. 2 and is the field due to a
point charge with charge +2Q.
-Q
a
+2Q
b
2Q
E  ke 2
r
Now find the field for a < r < b
The field must be zero inside a conductor in equilibrium. Thus from
Gauss’s law Qin is zero. There is a + 2Q from the point charge so we
must have Qa = -2Q on the inner surface of the spherical shell. Since the
net charge on the shell is -Q we can get the charge on the outer surface
from Qnet = Qa + Qb.
Qb= Qnet - Qa = -Q - (-2Q) = + Q.
Worked Example 5 cont’d
Find the field for r > b
From the symmetry of the problem, the field
in this region is radial and everywhere
perpendicular to the spherical Gaussian
surface. Furthermore, the field has the same
value at every point on the Gaussian surface
so the solution then proceeds exactly as in
Ex. 2, but Qin=2Q-Q.
-Q
a
+2Q
b

E  dA 

E dA  E

2
dA

E
4

r

Gauss’s law now gives:
E  4 r
2
 
Qin
0

2Q  Q
0

Q
0
1
Q
Q
or E 
 ke 2
2
4 0 r
r
