Transcript Gauss’s Law and Electric Potential

Gauss’s Law:
Gauss’ law relates the electric fields at points on
a (closed) Gaussian surface to the net charge
enclosed by that surface.
Gaussian surface, can have ANY shape, but the
shape that minimizes our calculations of
the electric field is one that mimics the
symmetry of the charge distribution.
The electric flux through a Gaussian surface is proportional to the
net number of electric field lines passing through that surface.
a
E0
What is the Flux of the field
through the butterfly net ?
a.
E0pa2
b.
0
c.
-E0pa2
d.
Not enough Info
A charged point particle is placed at the center of
a spherical Gaussian surface. The electric flux Φ is
changed if:
A) the sphere is replaced by a cube of the same volume
B) the sphere is replaced by a cube of one-tenth the volume
C) the point charge is moved off center (but still inside the
original sphere)
D) the point charge is moved to just outside the sphere
E) a second point charge is placed just outside the sphere
Electric Potential – Start by thinking about Potential Energy!!
Calculating the Potential from the Field:
Since the electrostatic force is conservative, all
paths yield the same result.
If we set potential Vi =0, then
One Example – a Line of Charge:
If l is the charge per unit length, then the
charge on length dx is:
Calculating the Field from the Potential:
Suppose that a positive test charge q0 moves
through a displacement from one equipotential
surface to the adjacent surface. The work the
electric field does on the test charge during the
move is W=-q0 dV. But we already said that the
work is a scalar product of :
Therefore,
That is,
Since E cos q is the component of E in the
direction of ds,
Therefore, the component of E in any
direction is the negative of the rate at
which the electric potential changes
with distance in that direction.
f
DU
DV = V f - Vi º
= - ò E • ds
q
i

V is
◦ Scalar
◦ Relative quantity
◦ Units: J/C = Volt

And it


E  V
◦ increases towards a + charge
◦ decreases towards a – charge
Equipotential Surfaces
Van de Graaff Generator
Strong E-fields are used to charge smoke and dust
particles, which are then attracted to the walls and fall as
ash precipitate.
Physics 117 – Fall 2012
For this next summer you have an internship with LA Department of Water and
Power. They have asked you to help design the air cleaners that will be used
on a new coal burning power plant. Fly ash, which is very light (typically 1 *
10-4g) and small in diameter (typically 1mm), exits the boiler along with the
hot gases. It is this fly ash with which you are concerned. Current plants are
using electrostatic precipitators to remove the fly ash. They wish to continue
using this technology as it removes over 99% of these particles, but you must
redesign the system to match the specifications of the new plant.
An ash precipitator removes solid emissions from smokestacks by electrostatic
means. As the gas, laden with fly ash, leaves the boiler it is sent through a
system that first gives the particles a net negative charge of 10-10C/particle.
The gas then proceeds to a vertical stack. There the small particles move
upward with the gases at 1m/s. In the stack, two oppositely charged parallel,
vertical plates are placed so that the smoke passes through them. The plates
are 15 meters tall and there is 1 meter between the plates. The horizontal
electric field between the plates detects the ash and traps it on one of the
plates. You want to know how powerful the electric field needs to be such
that most of the particles deflect into the plates and never make it out of the
stack. If the plates are square, what charge must each plate have on it to
produce that electric field? What voltage is required on a plate to produce
this electric field?
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which
feels the larger electric force?
1) proton
2) electron
3) both feel the same force
4) neither – there is no force
5) they feel the same
magnitude force but opposite
direction
Since F = qE and the proton and electron
have the same charge in magnitude, they
both experience the same force.
However, the forces point in opposite
directions because the proton and
electron are oppositely charged.
electron
electron
-
+
proton
proton

E
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. Which has
the larger acceleration?
1) proton
2) electron
3) both feel the same acceleration
4) neither – there is no acceleration
5) they feel the same magnitude
acceleration but opposite
direction
Since F = ma and the electron is much less
massive than the proton, then the electron
experiences the larger acceleration.
electron
electron
-
+
proton
proton

E
A proton and an electron are
in a constant electric field
created by oppositely
charged plates. You release
the proton from the positive
side and the electron from
the negative side. When it
strikes the opposite plate,
which one has more KE?
1) proton
2) electron
3) both acquire the same KE
4) neither – there is no change of
KE
5) they both acquire the same KE
but with opposite signs
Since PE = qV and the proton and electron
have the same charge in magnitude, they
both have the same electric potential energy
initially. Because energy is conserved, they
both must have the same kinetic energy
after they reach the opposite plate.
electron
electron
-
+
proton
proton

E
Four point charges are
arranged at the corners of
a square. Find the
electric field E and the
potential V at the center
of the square.
1) E = 0
V=0
2) E = 0
V0
3) E  0
V0
4) E  0
V=0
5) E = V regardless of the value
The potential is zero: the scalar
contributions from the two positive
charges cancel the two minus charges.
However, the contributions from the
electric field add up as vectors, and
they do not cancel (so it is non-zero).
Follow-up: What is the
direction of the electric field at
-Q
+Q
-Q
+Q
1
At which point
does V = 0?
5) all of them
2
+Q
3
–Q
4
All of the points are equidistant from both charges.
Since the charges are equal and opposite, their
contributions to the potential cancel out everywhere
along the mid-plane between the charges.
1) A and C
Which two points
have the same
potential?
2) B and E
3) B and D
4) C and E
5) no pair
Since the potential of a point charge is:
A
Q
V k
r
only points that are at the same distance
from charge Q are at the same potential.
This is true for points C and E.
They lie on an Equipotential Surface.
C
B
E
Q
D
Which requires zero
work, to move a positive
charge from P to points
1, 2, 3 or 4 ? All points
are the same distance
from P.
For path #3, you are moving in a
direction perpendicular to the field
lines. This means you are moving
along an equipotential, which
requires no work (by definition).
Follow-up: Which path requires the least
work?
1) P  1
2) P  2
3) P  3
4) P  4
5) all require the same
amount of work
3
2
1
P

E
4
24.4.5. Consider the equipotential
lines shown in the box. The labeled
cases indicate electric field line
drawings. Which of these cases best
matches the equipotential lines
shown?
a) 1
b) 2
c) 3
d) 4
e) None of these cases match the
equipotential lines shown.
24.9.1. Consider two conducting spheres with one having a larger radius
than the other. Both spheres carry the same amount of excess charge.
Which one of the following statements concerning the potential energy of
the two spheres is true?
a) The potential energy of the larger sphere is greater than that of the
smaller sphere.
b) The potential energy of the larger sphere is the same as that of the
smaller sphere.
c) The potential energy of the larger sphere is less than that of the smaller
sphere.