1 lecture 4 electric potential and/ potential energy ch. 25 review from lecture 3 cartoon - there is...
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Lecture 4 Electric Potential and/ Potential Energy Ch. 25
•Review from Lecture 3•Cartoon - There is an electric energy associated with the position of a charge.•Opening Demo - •Warm-up problems •Physlet •Topics
•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 filed from the potential•Electric potential energy from a system of point charge•Equipotential Surface•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
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Potential Energy and 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.
• = - W = -Work done by the electric force =
• Since , U = and
• Electric Potential difference = Potential energy change/ unit charge
• (independent of path, ds)
−q0 E ⋅dsa
b
∫
∫ ⋅−b
a
dsFab UUU −=
F =q0E
V =ΔU
q0
∫ ⋅−=−= dsEVVV ab
SI unit of electric potential is volt (V):1 Volt = 1 Joule/Coulomb (1 V = 1 J/C)
• Joule is too large a unit of energy when working at the atomic or molecular level, so use the electron-volt (eV), the energy obtained when an electron moves through a potential difference of 1 V. 1 eV = 1.6 x 10-19 J
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(independent of path, ds)
Therefore, electric force is a conservative force.
= - Work done by the electric force =
€
− F ⋅dsi
f
∫U =U f −Ui
qUV =
∫ ⋅−=−= dsEVVV if
y
x
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•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.
• 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
= −
rF
q0
⋅drs∫ = −
rE ⋅d
rs∫
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Example of finding the potential difference in a
Uniform FieldWhat is the electric potential difference for a unit positive charge moving in an uniform electric field from a to b?
E
Ed
a bx direction
)( ab
b
a
b
a
xxEdxEdsEV −−=−=⋅−= ∫∫d
EdV −=
VqU =
qEdU −=
dV =−EdxE =−dV / dx
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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
VVVqUV ab )90( −=−==
Potential energy is lower by
qU 9−=
9 μJ
CV 6101)9( −××−=
JoulesU 6109 −×−=
U = − 9 microJoules
= − 9 μ J
We also assumed that the potential at b was 0
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Example of a proton accelerated in a uniform fieldA 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 b
+
E = 105 V/m
d = 10 cm mqEd
v2=
v =2×1.6 ×10−19C ×105 V
m×0.1m1.67 ×10−27kg
v =1.4 ×108 ms
EdVVV ab −=−=
K = qEd
UK −=
U = qΔV = −qEd
0=+ KU
1
2mv2 =qEd
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What is the electric potential when moving from one point to another in a field due to a point charge?
V = −
rE ⋅d
rr∫
rE =
kqr2 r
Vf −Vi =−
rE ⋅d
rr
i
f
∫
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Vf −Vi =−
rE ⋅dr
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 ⋅dr
i
f
∫
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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)
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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
m
CC
Nm
Rkq
V10
192
29
10529.
106.11099.8
−
−
×
××⎟⎠⎞⎜
⎝⎛ ×
==
V =27.2JC
=27.2Volts
What is the electric potential energy of the electron at that point?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 volts.
Total energy of the electron in the ground state of hydrogen is - 13.6 eV
Also U= 2E = -27.2 eV. This agrees with above formula.
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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
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Potential due to a dipole
For two point charges, the total potential is the sum of the potentials of each point charge.
batotaldipole VVVV ,So +==
⎟⎠
⎞⎜⎝
⎛ −+=+=ba
badipole r)q(
rq
kVVV
⎟⎠
⎞⎜⎝
⎛ −=
ba
ab
rrrr
kq
We are interested in the regime where r>>d.
Vdipole ; kq
d cosθr2 ;
kpcosθr2
where p is the dipole moment.
As in fig 2, ra and rb are nearly parallel. And the difference in their length is dcosθ Also because r>>d, ra rb is approximately r2.
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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
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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
⎛
⎝⎜
⎞
⎠⎟
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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
V = −
rE ⋅d
rs∫
dV =−rE ⋅d
rs
dzdVE
dydVE
dxdVE
z
y
x
−=
−=
−=
dzEdyEdxEdV zyx −−−= d
rs =idx+ jdy+ kdz
rE =Exi + Ey j + Ezk
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Equipotential Surfaces
• Three examples
• 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?
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c) Uniform E field
V = constant in y and z directions
b) Point charge
(concentric shells)
a) 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
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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
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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
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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
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A metal slab is put in a uniform electric field of 106 N/C with the field perpendicular to both surfaces.
– 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. ∫ =⋅
0εinq
daE
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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 due to the charge on this ring at point P given by
Q))'R(z(
kV22 +
=
The potential dV at a point P due to the charged ring of radius R’ is
))'R(z(
'dR'R2k
))'R(z(
kdqdV
2222 +π=
+=
Integrating R’ from R’=0 to R’=R
∫+π=
R
022 ))'R(z(
'dR'R2kV )zRz(k2V 22 −+π=⇒