Transcript Lecture 4 Electric potential

Lecture 4 Electric Potential/Energy Chp. 25
•Cartoon - There is an electric energy associated with the position of a charge.
•Opening Demo •Warm-up problem
•Physlet
•Topics
•Electric potential energy and electric potential
•Equipotential Surface
•Calculation of potential from field
•Potential from a point charge
•Potential due to a group of point charges, electric dipole
•Potential due to continuous charged distributions
•Calculating the filed from the potential
•Electric potential energy from a system of point charge
•Potential of a charged isolated conductor
•Demos
•teflon and silk
•Charge Tester, non-spherical conductor, compare charge density at
Radii
•Van de Graaff generator with pointed objects
Electric potential
• The electric force is mathematically the same as gravity so it too must be a
conservative force. We will find it useful to define a potential energy as is the
case for gravity. Recall that the change in the potential energy in moving
from one point a to point b is the negative of the work done by the electric
force.
b
• U = Ub -Ua = - Work done by the electric force =   F  ds
b
• Since F=qE,
 U = q  E  ds and
a
a
• Electric Potential difference = Potential energy change/ unit charge
•  V= =  U/q
•
V = Vb -Va = -  E.ds (independent of path ds)
f
U = Ub -Ua = - Work done by the electric force =
  F  ds
i

 V= =  U/q
V = Vf -Vi = -  E.ds (independent of path ds)
Therefore, electric force is a conservative force
V = Vf -Vi = -  E.ds (independent of path ds)
•The SI units of V are Joules/ Coulomb or Volt. E is
N/C or V/m.
•The potential difference is the negative of the work done
per unit charge by an electric field on a positive unit charge
when it moves from point a to point b.
•We are free to choose V to be 0 at any location. Normally
V is chosen to be 0 at infinity for a point charge.
Examples
•
What is the electric potential difference for a positive charge moving in an
uniform electric field?
E
E
d
x direction
a
b
b
b
a
a
V    E  ds   E  dx   E ( xb  xa)
V = -Ed
U = q V
U = -qEd
d
Examples
•
In a 9 volt battery, typically used in IC circuits, the positive terminal has a
potential 9 v higher than the negative terminal. If one micro-Coulomb of
positive charge flows through an external circuit from the positive to negative
terminal, how much has its potential energy been changed?
V 
q
U
 Vb  Va  0  9
q
U  9q
= -9V x 10-6 C
U = -9V x 10-6 Joules
U = -9 microJoules
Potential energy is lower by 9 microJoules
Examples
A proton is placed in an electric field of E=105 V/m and released. After going 10
cm, what is its speed?
Use conservation of energy.
a
E = 105 V/m
d = 10 cm
b
v
+
V = Vb – Va = -Ed
v
U = q V
U + K = 0
K = (1/2)mv2
K = -U
(1/2)mv2 = -q V = +qEd
2qEd
m
2 1.6 1019 C 105 Vm 1m
1.67  10 23 kg
v  1.4 106
m
s
Electric potential due to point charges
• Electric field for a point charge is E = (kq/r2). The change in electric
potential a distance dr away is dV = - E.dr.
•
Integrate to get
V = -  kq/r2 dr = kq/r + constant.
• We choose the constant so that V=0 at r = .
• Then we have V= kq/r for point charge.
• V is a scalar, not a vector
• V is positive for positive charges, negative for negative charges.
• r is always positive.
• For many point charges, the potential at a point in space is the sum
V=  kqi/ri.
y
1
r1
p
2
r2
r3
3
x
 q1 q 2 q3 
V  k   
 r1 r 2 r 3 
f
V f Vi    E  ds
i

V f Vi    E  dr  kq 
R



q
0 Vi  k
R
V 
k
kq
R
1
4 0
Replace R with r
V
1
q
4 0 r
eqn 25-26
E
kq
r2

1
1
dr

kq
r2
rR
Electric potential for a positive point charge
V (r)  kq/r
r x y
2
2
Electric potential due to a positive point charge
Hydrogen atom.
•What is the electric potential at a distance of 0.529 A from
the proton?
+
r
r = 0.529 A
• V = kq/r = 8.99*1010 N m2//C2 *1.6*10-19 C/0.529*10-10m
• V = 27. 2 J/C = 27. 2 Volts
Ways of Finding V
•
•
Direct integration. Since V is a scalar, it is easier to evaluate V than E.
Find V on the axis of a ring of total charge Q. Use the formula for a point charge,
but replace q with elemental charge dq and integrate.
Point charge V = kq/r.
For an element of charge dq, dV = kdq/r.
r is a constant as we integrate.
V =  kdq/r
=  kdq/(z2+R2)1/2
= k/(z2+R2)1/2  dq
= k/(z2+R2)1/2 Q
This is simpler than finding E because V is not a vector.
V = kq/r
More Ways of finding V and E
•Use Gauss’ Law to find E, then use V= -  E.ds to get V
–Suppose Ey = 1000 V/m. What is V? V = -1000 y
(Uniform field)
V = Ed
•More generally,
If we know V, how do we find E? dV= - E. ds
–ds = i dx +j dy+k dz
and dV = - Exdx - Eydy - Ezdz
Ex = - dV/dx,
Ey = - dV/dy,
Ez = - dV/dz.
–So the x component of E is the derivative of V with respect to x, etc.
–If Ex = 0, then V = constant in that direction. Then lines or surfaces
–on which V remains constant are called equipotential lines or surfaces.
–See examples
Equipotential Surfaces
• Example: For a point charge, find the equipotential surfaces. First
draw the field lines, then find a surface perpendicular to these
lines. See slide.
• What is the equipotential surface for a uniform field in the y
direction? See slide.
• What is the obvious equipotential surface and equipotential
volume for an arbitrary shaped charged conductor?
• See physlet 9.3.2 Which equipotential surfaces fit the field lines?
a)
c)
Two point charges
b)
Point charge
Uniform E field
E = Ex , Ey = 0 , Ez = 0
Ex = dv/dx
(ellipsoidal concentric shells)
(concentric shells)
V = Ex d
V = constant in y and z
directions
How does a conductor shield the interior from an
exterior electric field?
•Start out with a uniform electric field
with no excess charge on conductor.
Electrons on surface of conductor adjust
so that:
1. E=0 inside conductor
2. Electric field lines are perpendicular
to the surface. Suppose they weren’t?
3. Does E = s/0 just outside the conductor
4. Is suniform over the surface?
5. Is the surface an equipotential?
6. If the surface had an excess charge, how would your answers change?
Dielectric Breakdown: Application of Gauss’s Law
• If the electric field in a gas exceeds a certain value, the gas breaks down
and you get a spark or lightning bolt if the gas is air. In dry air at
STP, you get a spark when E = 3*106 V/m. To examine this we model
the shape of a conductor with two different spheres at each end:
V = constant on
surface of conductor
+
+
+
+
+
+
+
+
+
+
1
2
Radius R
•The surface is at the same potential everywhere, but charge density and
electric fields are different. For a sphere, V= q/(4 0 r) and q = 4r2s,
then V = (s/ 0 )*r. Since E = s/ 0 near the surface of the conductor,
we get V=E*r. Since V is a constant, E must vary as 1/r and s as 1/r.
Hence, for surfaces where the radius is smaller, the electric field and charge
will be larger. This explains why:
V = constant on
surface of conductor
+
+
+
+
+
+
+
+
+
+
1
2
Radius R
This explains why:
•
Sharp points on conductors have the highest electric fields and
cause corona discharge or sparks.
•
Pick up the most charge with charge tester from the pointy regions
of the non-spherical conductor.
•
Use non-spherical metal conductor charged with teflon rod. Show
variation of charge across surface with charge tester.
V = constant on
surface of conductor
+
+
+
+
+
+
+
+
+
+
1
2
Radius R
Cloud
+
+
+
+
Van de Graaff
Show lightning rod demo
with Van de Graaff
-
-
-
-
A metal slab is put in a uniform electric field of 106 N/C
with the field perpendicular to both surfaces
.
– Draw the appropriate model for the problem.
– Show how the charges are distributed on the
conductor.
– Draw the appropriate pill boxes.
– What is the charge density on each face of
the slab?
– Apply Gauss’s Law.  E.dA = qin/0
Slab of metal In a uniform Electric field
E = 106 N/C
Slab of metal
In a uniform Electric
field
-
+
+
+
+
Gaussian
Pill Box
Evaluate E.dA = qin/0
• Left side of slab - E*A + 0*A = As/0,
E = - s /0, s = - 106 N/C *10-11 C2/Nm2 = -10-5 N/m2
.Right side of slab
0*A + E*A = As/0,
E = s /0, s = 106 N/C *10-11 C2/Nm2 = +10-5 N/m2
(note that charges arranges themselves so that field
inside is 0)
What is the electric potential of a
uniformly charged circular disk?
Warm-up set 4
1. HRW6 25.TB.37. [119743] The equipotential surfaces associated with an isolated point charge are:
concentric cylinders with the charge on the axis
vertical planes
horizontal planes
radially outward from the charge
concentric spheres centered at the charge
2. HRW6 25.TB.17. [119723] Two large parallel conducting plates are separated by a distance d,
placed in a vacuum, and connected to a source of potential difference V. An oxygen ion, with charge 2e,
starts from rest on the surface of one plate and accelerates to the other. If e denotes the magnitude of the
electron charge, the final kinetic energy of this ion is:
2eV
eVd
Vd / e
eV / d
eV / 2
3. HRW6 25.TB.10. [119716] During a lightning discharge, 30 C of charge move through a
potential difference of 1.0 x 108 V in 2.0 x 10-2 s. The energy released by this lightning bolt is:
1.5 x 1011 J
1500 J
3.0 x 109 J
3.3 x 106 J
6.0 x 107 J