lecture 4 work, electric potential and potential energy ch. 25
DESCRIPTION
Lecture 4 Work, Electric Potential and 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 - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/1.jpg)
1
Lecture 4 Work, Electric Potential and 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
![Page 2: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/2.jpg)
2
W =
rF ⋅d
rs
i
f
∫ = qrE ⋅d
rs
i
f
∫
ΔU = −W
![Page 3: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/3.jpg)
3
ΔV =ΔU
q
![Page 4: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/4.jpg)
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 is independent 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 case for gravity.
• And the electric potential W =
rF ⋅d
rs
i
f
∫ = qrE ⋅d
rs
i
f
∫
ΔV =ΔU
q
ΔU = −W
![Page 5: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/5.jpg)
5
Lets start with a uniform electric field and find the work done for a positive test charge.
b
a
E
P2 P3
P1
c
W = q
rE ⋅drs
i
f
∫ =qEcosθ s
![Page 6: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/6.jpg)
6
Find work done along path W12 for a positive test charge
b
a
E
P2 P3
P1
c
F
W12 =0W12 =qEcos90 a
W = q
rE ⋅drs
i
f
∫ =qEcosθ s
![Page 7: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/7.jpg)
7
Find Work along path W23
b
a
E
P2 P3
P1
c
F
W23 =qEb
W = q
rE ⋅drs
i
f
∫ =qEcosθ s
W23 =qEcos0 b
![Page 8: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/8.jpg)
8
W12 + W23 = 0 + qEb =qEb
Compare this work done along path W13
![Page 9: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/9.jpg)
9
Work done along path W13
b
a
P2 P3
P1
c
Fθ
W13 =qEcosθ c=qE
bc
c=qEb
W = q
rE ⋅drs
i
f
∫ =qEcosθ s
θ
![Page 10: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/10.jpg)
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 is true for any field that is a only a function of r and is along r.• It only depends on the end points i and f.
•This means we can define a function at every point in space and when we take the difference of that function between any two points, it is equal to the negative of the work done.
![Page 11: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/11.jpg)
11
b
a
P2 P3
P1
c
Fθ
When we go from P1 to P3 we evaluate the Work function at P3 and subtract the value at P1 and then the a difference equals the negative of the work done in going form P1 to P3. This function is called the potential energy function
![Page 12: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/12.jpg)
12
Example of finding the Potential Energy Function U in a Uniform Field
What is the electric potential difference for a unit positive charge moving in an uniform electric field from a to b?
E
Ed
a b
ΔU =Ub −Ua = −W = − q
rE ⋅d
rs
a
b
∫ = −Eq dxa
b
∫ = −Eq(xb − xa)
ab UUU −=Δ
If we set the origin at xb = 0, and measure from b to a, then
ΔU = qEd
This is analogy with gravitation where we U =mgh.
Ub=0 and the potential energy function is U=qEd
![Page 13: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/13.jpg)
13
Now define the Electric Potential Difference ΔV which does not depend on charge.
ΔV =ΔU
q
ΔU = −W
ΔV =−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.
![Page 14: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/14.jpg)
14
ΔV = Ed
ΔV =ΔU
q
ΔV =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.
ΔU = qEd
Find the potential difference ΔV for a uniform electric field
![Page 15: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/15.jpg)
15
Note relationship between potential 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.
ΔV =
−W
q0
= −
rF
q0
⋅drs∫ = −
rE ⋅d
rs∫ dV =−Edx
E =−dV / dx
![Page 16: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/16.jpg)
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 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?
q
![Page 17: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/17.jpg)
17
Generalize concept of electric potential energy and potential
difference for any electric field
![Page 18: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/18.jpg)
18
(independent of path, ds)
Therefore, electric force is a conservative force.
= - Work done by the electric force = −
rF ⋅d
rs
i
f
∫ΔU =U f −Ui
qUV Δ=Δ
ΔV = Vf −Vi = −
rE ⋅d
rs∫
y
x
![Page 19: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/19.jpg)
19
Find the electric potential when moving from one point to another in a field due to a point charge?
ΔV = −
rE ⋅d
rr∫
rE =
kqr2 r
Vf −Vi =−
rE ⋅d
rr
i
f
∫
![Page 20: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/20.jpg)
20
Vf −Vi =−
rE ⋅d
rr
R
∞
∫ =−kqcos0o 1r2 d
R
∞
∫ r =kq1r R
∞
=kq(1∞−1R)
R
kqV =
04
1
πε=k
eqn 25-26
Replace R with r
V =1
4πε 0
qr
Vf −Vi =0−Vi =−kqR
Potential of a point charge at a distance R
Vf −Vi =−
rE ⋅d
rr
i
f
∫
![Page 21: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/21.jpg)
21
Electric potential for a positive point charge
V (r) =kqr
r = x2 + y2
• 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)
![Page 22: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/22.jpg)
22
Hydrogen atom.
• What is the electric potential at a distance of 0.529 A from the 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 =kqR
=8.99 ×109 Nm2
C2( )×1.6 ×10−19C
.529 ×10−10m
V =27.2JC
=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 eV
Also U= 2E = -27.2 eV. This agrees with above formula.
![Page 23: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/23.jpg)
23
What is the electric potential due to several point charges?
For many point charges, the potential at a point in space is the 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
![Page 24: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/24.jpg)
24
Four Point charges
![Page 25: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/25.jpg)
25
What is the potential due to a dipole?
Two point charges that are opposite and equal
![Page 26: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/26.jpg)
26
Potential for a Continuous Distribution of Charge
Point charge
For an element of charge
Integrate
rkq
V =
rkdq
dVdq = ,
∫=r
kdqV
![Page 27: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/27.jpg)
27
Chaper 24 Problem 22. With V = 0 at infinity, what is the electric potential at P, the center of curvature of the uniformly charged nonconducting rod?
![Page 28: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/28.jpg)
28
Chapter 24 Problem 26. What is the magnitude of the net electric potential 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 radius 4.0 cm, subtending an angle of 90° about the center of the full circle3. An electric dipole with a dipole moment that is perpendicular to a
radial line and that has magnitude 1.28 multiplied by 10-21 C·m
![Page 29: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/29.jpg)
29
Figure 24-44 shows a thin plastic rod of length L and uniform positive charge Q lying on an x axis. With V = 0 at infinity, find the electric potential at point P1 on the axis, at distance d from one end of the rod.
![Page 30: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/30.jpg)
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 and integrate.
Point charge
For an element of charge
r is a constant as we integrate.
This is simpler than finding E because V is not a vector.
rkq
V =
rkdq
dVdq = ,
∫=r
kdqV
Q)Rz(
kV22 +
=⇒∫+
= dqRz
k
)( 22
∫+
=)( 22 Rz
kdq
![Page 31: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/31.jpg)
31
Potential due to a line charge
rdq
kdV =We know that for an element of charge dq
the potential is
rdxkVd ,So λ=
For the line charge let the charge density be λ. Then dq=λdx
22 dxr ,But +=
22 dx
dxkVd ,Then+
λ=
Now, we can find the total potential V produced by the rod at point P by integrating along the length of the rod from x=0 to x=L
∫∫∫+
λ=+
λ==L
022
L
022
L
0 dx
dxkdx
dxk VdV ⇒ V = kλ ln(x + x2 + d 2 )0
L
)dln)dLL(ln(k Vo,S 22 −++λ= or, V =kλ lnL + L2 +d2
d
⎛
⎝⎜
⎞
⎠⎟
![Page 32: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/32.jpg)
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 V remains constant are called equipotential lines or surfaces.–See example on next slide
ΔV = −
rE ⋅d
rs∫
dV =−rE ⋅d
rs
dzdVE
dydVE
dxdVE
z
y
x
−=
−=
−=
dzEdyEdxEdV zyx −−−=
drs =idx+ jdy+ kdz
rE =Exi + Ey j + Ezk
![Page 33: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/33.jpg)
33
Now find the electric field at point P1 on the axis, at distance d from one end of the rod. Find the x and y components Ex and Ey.
![Page 34: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/34.jpg)
34
Equipotential Surfaces
• Three examples
• What is the equipotential surface and equipotential volume for an arbitrary shaped charged conductor?
• See physlet 9.3.2 Which equipotential surfaces fit the field lines?
![Page 35: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/35.jpg)
35
a) Uniform E field
V = constant in y and z directions
b) Point charge
(concentric shells)
c) Electric Dipole
(ellipsoidal concentric shells)
E =Ex,Ey =0,Ez =0
Ex =−dVdx
V =−Exd
Blue lines are the electric field lines
Orange dotted lines represent the equipotential surfaces
x
![Page 36: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/36.jpg)
36
Electric Potential Energy U of a system of charges
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(−qqa
+qq2a
−qqa
−qqa
+qq2a
−qqa)
U =k(−4q2
a+2q2
2a)
![Page 37: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/37.jpg)
37
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
![Page 38: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/38.jpg)
38
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 ≥3×104 Vcm
12
V = constant on surface of conductor Radius r2
r1
![Page 39: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/39.jpg)
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 the non-spherical conductor.
• Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester.
Van de Graaff
+ + + +
- - - -
Radius R
1
2
V = constant on surface of conductor
Cloud
![Page 40: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/40.jpg)
40
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 = just outside the conductor
4. Is uniform over the surface?
5. Is the surface an equipotential?
6. If the surface had an excess charge, how would your answers change?
0ε
![Page 41: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/41.jpg)
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 an area of 2πR’dR’ and it’s charge is dq = dA = 2πR’)dR’ where =Q/(πR2), the surface charge density. The potential dV at point P due to the charge dq on this ring given by
Integrating R’ from R’=0 to R’=R
q
dV =kdq
(z2 + (R')2 )
dV =k2πR'dR'
(z2 _(R')2 )⇒
V =k2πR'dR'
(z2 + (R')2 )0
R
∫
⇒ V = 2kσπ ( z2 + R2 − z)
![Page 42: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/42.jpg)
42
Chapter 24 Problem 19
The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.31 D, where 1 D = 1 debye unit = 3.34 multiplied by 10-30 C·m. Calculate the electric potential due to an ammonia molecule at a point 44.0 nm away along the axis of the dipole. (Set V = 0 at infinity.)
![Page 43: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/43.jpg)
43
Chapter 24 Problem 55
Two metal spheres, each of radius 1.0 cm, have a center-to-center separation 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 the separation is large enough for us to assume that the charge on each sphere 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.
![Page 44: Lecture 4 Work, Electric Potential and Potential Energy Ch. 25](https://reader035.vdocument.in/reader035/viewer/2022062517/56813e5e550346895da8605c/html5/thumbnails/44.jpg)
44
Chapter 24 Problem 57
A metal sphere of radius 11 cm has a net charge of 2.0 multiplied by 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's surface?(c) At what distance from the sphere's surface has the electric potential decreased by 500 V?