Download Brief recap Direction of Electric Field Problem 1

PHY2054 Announcements
Brief recap
1. Discussion section quizzes begin this week!
• Chapter 15.1 – 15.4
„
2. First WebAssign homework due this Friday,
11 pm
3. We will skip Section 15.7
Positive (+) and negative (-) charges
„
„
„
Charging by conduction or induction
„
Coulomb’s Law is used to compute
electrical forces on charges
Questions?
„
„
„
Positive charge:
E is directed
away from the
charge
Negative charge:
E is directed
toward the
charge
Vector law!
Electrical fields communicate the
presences of other charges
„
Direction of Electric Field
Like charges repel and unlike charges
attract one another
Fundamental charge is e = 1.6 x 10-19 C
More fundamental than electrical forces
Electric field is a vector quantity
Coulomb’s Law
F = ke
q1 q2
r2
Electric Field
E = ke
Q
r2
(for point charges!)
Problem 1
Find the point on the x-axis where the
electric field is zero.
q1 = +1.0 x 10-6 C
q2 = +5.0 x 10-6 C
q1
r12
q2
x
r12 = 10 cm
1
15.5 Electric Field Lines
Positive Point Charge
„
„
Electric Field Line Patterns
Field lines
go radially
outward
A convenient way to visualize
electric field patterns
„
Drawing field lines - point in
the direction of the field vector
at any point
Negative Point Charge
„
„
The field lines are related to
the field as follows:
„
„
Field lines
go radially
inward
The electric field vector E is
tangent to the electric field
lines at each point
The number of lines per unit
area through a surface
perpendicular to the lines is
proportional to strength of the
electric field
Electric Field Line Patterns
„
„
An electric dipole
consists of two
equal and
opposite charges
The high density
of lines between
the charges
indicates the
strong electric
field in this region
E
An electric dipole
consists of two
equal and
opposite charges
The high density
of lines between
the charges
indicates the
strong electric
field in this region
Electric Field Line Patterns
„
„
„
„
Example: two equal but like
point charges
The bulging out of the field
lines between the charges
indicates the repulsion
between the charges
The low field lines between
the charges indicates a weak
field in this region
At a great distance from the
charges, the field would be
approximately that of a single
charge of 2q
2
Rules for Drawing Electric
Field Lines
„
The lines for a group of charges must
begin on positive charges and end on
negative charges
„
„
„
15.6 Conductors in
Electrostatic Equilibrium
„
„
For excess of charge, some lines will begin
or end infinitely far away
„
The number of lines drawn leaving a
positive charge or ending on a negative
charge is proportional to the magnitude
of the charge
No two field lines can cross each other
„
„
„
E just outside a charged
conductor is perpendicular to the
conductor’s surface
„
„
„
Consider what would
happen it this was
not true
The component
along the surface
would cause the
charge to move
It would not be in
equilibrium
When no net motion of charge occurs within a
conductor, the conductor is said to be in
electrostatic equilibrium
An isolated conductor has the following
properties:
The electric field is zero everywhere inside the
conducting material
Any excess charge on an isolated conductor resides
entirely on its surface
The electric field just outside a charged conductor is
perpendicular to the conductor’s surface
On an irregularly shaped conductor, the charge
accumulates at locations where the radius of
curvature of the surface is smallest (that is, at sharp
points)
Electric Flux
„
„
„
Field lines penetrating an area A
perpendicular to the field
The product of EA is the electric
flux, ΦE
In general:
„
„
ΦE = E A cos θ
The perpendicular to the area
A is at an angle θ to the field
For closed surfaces, the flux
lines passing into the interior
are negative; those passing
out of the interior are positive
3
Simple Example:
Electric Field of a Charged Thin
Spherical Shell
Gauss’ Law
„
Gauss’ Law states that the electric flux
through any closed surface is equal to the
net charge Q inside the surface divided by
εo
ΦE =
Qinside
εo
„
„
εo is the permittivity of free space and equals
8.85 x 10-12 C2/Nm2
„
„
The calculation of the field outside the shell is
identical to that of a point charge
E=
Coulomb constant: ke = 1/4πεo
A Gaussian surface does not have to coincide
with the surface of a physical object
Problem 2
„
r1 = 10 cm
Find:
a) The charge on the inner and
outer surfaces of the shell
b) The distance r2
q1
„
q2
„
r2
Use a cylindrical Gaussian
surface
The flux through the ends
is EA
„
r1
The electric field inside the shell is zero
Electric Field of a Nonconducting
Plane Sheet of Charge
q1 = +3.0 x 10-6 C
q2 = -1.0 x 10-6 C
A solid conducting sphere is
located at the center of the
spherical conducting shell. The
sphere has a total charge q1 and
the shell has a total charge q2.
The ratio Er1/Er2 = 10
Q
Q
= ke 2
4πr 2 εo
r
„
no field through the curved
part of the surface
The total charge is Q = σA
E=
„
σ is the surface
charge density (C/m2)
σ
2ε 0
The field is uniform
4
Parallel Plate Capacitor
„
„
The device consists of
plates of positive and
negative charge
The total electric field
between the plates is
given by
E =
„
σ
εo
The field outside the
plates is zero
5
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