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Lecture 4 Work, Electric Potentialand Potential Energy Ch. 25
• Topics• Work, electric potential energy and electric potential• 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 electric field from a potential• Electric potential energy from a system of point charges• Equipotential Surface• Potential of a charged isolated conductor
• Demos• Teflon and silk• Charge Tester, non-spherical conductor, compare charge density at different radii.
•Elmo•Potential in the center of four charges•Potential of a electric dipole
• Polling
4
Work, Potential Energy and Electric potential• The electric force is mathematically the same as gravity so it too must
be a conservative force. We want to show that the work done isindependent of the path and only depends on the endpoints. Then
the force is said to be a conservative force.
• First start with work Work done by the electric force =
• Then we will find it useful to define a potential energy as is the casefor gravity.
• And the electric potential
W =!F !d!s
i
f
" = q!E !d!s
i
f
"
!V =!U
q
!U = "W
5
Lets start with a uniform electric field and findthe work done for a positive test charge.
b
a
E
P2 P3
P1
c
W = q!E !d!s
i
f
" = qE cos# s
6
Find work done along path W12 for a positive testcharge
b
a
E
P2 P3
P1
c
F
W12= 0W
12= qE cos90 a
W = q!E !d!s
i
f
" = qE cos# s
7
Find Work along path W23
b
a
E
P2 P3
P1
c
F
W
23= qEb
W = q!E !d!s
i
f
" = qE cos# s
W23= qE cos0 b
9
Work done along path W13
b
a
P2 P3
P1
c
Fθ
W13= qEcos! c = qE
b
cc = qEb
W = q!E !d!s
i
f
" = qE cos# s
θ
10
Conclusion• Work done along path W12 + W23 = W13.
• Work is independent of the particular path.• Although we proved it for a uniform field, it istrue for any field that is a only a function of r andis along r.• It only depends on the end points i and f.
•This means we can define a function at everypoint in space and when we take the difference ofthat function between any two points, it is equalto the negative of the work done.
11
b
a
P2 P3
P1
c
Fθ
When we go from P1 to P3 we evaluate the Work function atP3 and subtract the value at P1 and then the a differenceequals the negative of the work done in going form P1 to P3.This function is called the potential energy function
12
Example of finding the Potential EnergyFunction U in a Uniform Field
What is the electric potential difference for a unit positivecharge moving in an uniform electric field from a to b?
E
Ed
a b
!U =Ub "Ua = "W = " q!E #d!s
a
b
$ = "Eq dxa
b
$ = "Eq(xb " xa)
abUUU !="
If we set the origin at xb = 0, and measure from b to a, then
!U = qEd
This is analogy with gravitation where we U =mgh.
Ub=0 and the potential energy function is U=qEd
13
Now define the Electric Potential Difference ΔVwhich does not depend on charge.
!V =!U
q
!U = "W
!V ="W
q0
• 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 one point to another.
14
!V = Ed
!V =!U
q
!V =qEd
q
For a battery of potential difference of 9 volts you would say that the positive terminal is 9 volts above the negative terminal.
!U = qEd
Find the potential difference ΔV for a uniform electric field
15
Note relationship betweenpotential and electric field
• V is a scalar not a vector. Simplifies solving problems.
• We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or 0 at infinity for a point charge.
!V ="W
q0
= "
!F
q0
#d!s$ = "
!E #d!s$ dV = !Edx
E = !dV / dx
16
Example for a battery in a circuit• In a 9 volt battery, typically used in IC circuits, the
positive terminal has a potential 9 v higher than thenegative terminal. If one micro-Coulomb of positivecharge flows through an external circuit from the positiveto negative terminal, how much has its potential energybeen changed?
q
18
(independent of path, ds)
Therefore, electric force is a conservative force.
= - Work done by the electric force =
!!F "d!s
i
f
#!U =U f "Ui
q
UV
!=!
!V = Vf "Vi = "
!E #d!s$
y
x
19
Find the electric potential when moving fromone point to another in a field due to a pointcharge?
!V = "
!E #d!r$
!E =
kq
r2r
Vf !Vi = !!E "d!r
i
f
#
20
Vf !Vi = !!E "d!r
R
#
$ = !kqcos0"1
r2d
R
#
$ r = kq1
r R
#
= kq(1
#!1
R)
R
kqV =
04
1
!"=k
eqn 25-26
Replace R with r
V =1
4!" 0
q
r
Vf !Vi = 0 !Vi = !kq
R
Potential of a point charge at a distance R
Vf !Vi = !!E "d!r
i
f
#
21
Electric potential for a positive point charge
V (r) =kq
r
r = x2+ y
2
• V is a scalar• 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
simple algebraic sum (Not a vector sum)
22
Hydrogen atom.• What is the electric potential at a distance of 0.529 A fromthe proton? 1A= 10-10 m
Electric potential due to a positive point charge
r = 0.529 A
•What is the electric potential energy of the electron at that point?
V =kq
R=8.99 !10
9 Nm2
C2( ) !1.6 !10"19
C
.529 !10"10m
V = 27.2J
C= 27.2Volts
U = qV= (-1.6 x 10-19 C) (27.2 V) = - 43.52 x 10-19 Jor - 27.2 eV where eV stands for electron volt.Note that the total energy E of the electron in the ground state of hydrogen is - 13.6 eVAlso U= 2E = -27.2 eV. This agrees with above formula.
23
What is the electric potential due toseveral point charges?
For many point charges, the potential at a point in space isthe simple algebraic sum (Not a vector sum)
!"#
$%&
++=3
3
2
2
1
1
r
q
r
q
r
qkV
V =kqi
rii
!
r1
xr3
y
r2
q1q2
q3
26
Potential for a Continuous Distribution ofCharge
Point charge
For an element of charge
Integrate
r
kqV =
r
kdqdVdq = ,
!=r
kdqV
27
Chaper 24 Problem 22. With V = 0 at infinity, what is theelectric potential at P, the center of curvature of the uniformlycharged nonconducting rod?
28
Chapter 24 Problem 26. What is the magnitude of the net electricpotential at the center?1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm2. A second thin rod of charge 2.0 µC that forms a circular arc of radius4.0 cm, subtending an angle of 90° about the center of the full circle3. An electric dipole with a dipole moment that is perpendicular to aradial line and that has magnitude 1.28 multiplied by 10-21 C·m
29
Figure 24-44 shows a thin plastic rod of length L anduniform positive charge Q lying on an x axis. With V =0 at infinity, find the electric potential at point P1 on theaxis, at distance d from one end of the rod.
30
Potential due to a ring of charge• Direct integration. Since V is a scalar, it is easier to evaluate
V than E.• Find V on the axis of a ring of total charge Q. Use the formula
for a point charge, but replace q with elemental charge dq andintegrate.
Point charge
For an element of charge
r is a constant as we integrate.
This is simpler than finding E because Vis not a vector.
r
kqV =
r
kdqdVdq = ,
!=r
kdqV
Q)Rz(
kV
22+
=!!+
= dqRz
k
)( 22
!+
=
)( 22 Rz
kdq
31
Potential due to a line charge
r
dqkdV =
We know that for an element of charge dqthe potential is
r
dxkVd ,So!
=
For the line charge let the charge density be λ.Then dq=λdx
22dxr ,But +=
22dx
dxkVd ,Then
+
!=
Now, we can find the total potential V produced by the rod at point P byintegrating along the length of the rod from x=0 to x=L
!!!+
"=
+
"==
L
022
L
022
L
0 dx
dxk
dx
dxk VdV !V = k" ln(x + x
2+ d
2)0
L
)dln)dLL(ln(k Vo,S 22!++"= or, V = k! ln
L + L2+ d
2
d
"
#$
%
&'
32
A new method to find E if the potential is known. If we know V, how do we find E?
So the x component of E is the derivative of V with respect to x, etc.–If V = a constant, then Ex = 0. The lines or surfaces on which Vremains constant are called equipotential lines or surfaces.–See example on next slide
!V = "
!E #d!s$
dV = !!E "d!s
dz
dVE
dy
dVE
dx
dVE
z
y
x
!=
!=
!=
dzEdyEdxEdV zyx !!!=
d!s = idx + jdy + kdz
!E = Exi + Ey j + Ezk
33
Now find the electric field at point P1 on the axis, atdistance d from one end of the rod. Find the x and ycomponents Ex and Ey.
34
Equipotential Surfaces
• Three examples
• What is the equipotential surface and equipotentialvolume for an arbitrary shaped charged conductor?
• See physlet 9.3.2 Which equipotential surfaces fit thefield lines?
35
a) Uniform E field
V = constant in y and zdirections
b) Point charge
(concentric shells)
c) Electric Dipole
(ellipsoidal concentric shells)
E = Ex ,Ey = 0,Ez = 0
Ex = !dV
dx
V = !Exd
Blue lines are the electric field lines
Orange dotted lines represent the equipotential surfaces
x
36
Electric Potential Energy U of a system ofcharges
How much work is required to set up the arrangement of Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are initially infinitely far apart and at rest?
q1q2
q3q4
W = U
U = k(q1q2
r12
+q1q3
r13
+q1q4
r14
+q2q3
r23
+q2q4
r24
+q3q4
r34
)
U = k(!qq
a+qq
2a!qq
a!qq
a+qq
2a!qq
a)
U = k(!4q
2
a+2q
2
2a)
38
Dielectric Breakdown: Application of Gauss’s Law
If the electric field in a gas exceeds a certain value, the gasbreaks down and you get a spark or lightning bolt if the gasis air. In dry air at STP, you get a spark when
E ! 3"104V
cm
12
V = constant on surface of conductor Radius r2
r1
39
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 thenon-spherical conductor.
• Use non-spherical metal conductor charged with teflon rod. Showvariation of charge across surface with charge tester.
Van de Graaff+ + + +
- - - -
Radius R
1
2
V = constant on surface of conductor
Cloud
40
How does a conductor shield the interior from anexterior electric field?
• Start out with a uniform electric fieldwith no excess charge on conductor.Electrons on surface of conductor adjustso that:
1. E=0 inside conductor2. Electric field lines are perpendicularto the surface. Suppose they weren’t?3. Does E = just outside the conductor4. Is σ uniform over the surface?5. Is the surface an equipotential?
6. If the surface had an excess charge, how would your answers change?
0!
"
41
What is the electric potential of a uniformly charged circular disk?
We can treat the disk as a set of ring charges.The ring of radius R’ and thickness dR’ has anarea of 2πR’dR’ and it’s charge is dq = σdA = σ(2πR’)dR’ where σ=Q/(πR2), the surface chargedensity. The potential dV at point P due to thecharge dq on this ring given by
Integrating R’ from R’=0 to R’=R
q
dV =kdq
(z2+ (R ')
2)
dV =k!2"R 'dR '
(z2_(R ')
2)
!
V = k!2"R 'dR '
(z2+ (R ')
2)0
R
#
!V = 2k"# ( z2+ R
2$ z)
42
Chapter 24 Problem 19
The ammonia molecule NH3 has a permanent electric dipolemoment equal to 1.31 D, where 1 D = 1 debye unit = 3.34multiplied by 10-30 C·m. Calculate the electric potential due to anammonia molecule at a point 44.0 nm away along the axis of thedipole. (Set V = 0 at infinity.)
43
Chapter 24 Problem 55
Two metal spheres, each of radius 1.0 cm, have a center-to-centerseparation of 2.2 m. Sphere 1 has charge +2.0 multiplied by 10-8 C.Sphere 2 has charge of -3.8 multiplied by 10-8 C. Assume that theseparation is large enough for us to assume that the charge on eachsphere is uniformly distributed (the spheres do not affect each other).Take V = 0 at infinity.
(a) Calculate the potential at the point halfway between the centers.(b) Calculate the potential on the surface of sphere 1.(c) Calculate the potential on the surface of sphere 2.
44
Chapter 24 Problem 57
A metal sphere of radius 11 cm has a net charge of 2.0 multipliedby 10-8 C.
(a) What is the electric field at the sphere's surface?(b) If V = 0 at infinity, what is the electric potential at the sphere'ssurface?(c) At what distance from the sphere's surface has the electricpotential decreased by 500 V?