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University Physics (PHYS 2326)
Lecture 4
Electrostatics
Electric flux and Gauss’s law
Electrical energy
potential difference and
electric potential
potential energy of charged
conductors
Chapters 22-23
http://www2.swgc.mun.ca/physics/physlets/jepoten2.html
The Coulomb force is a conservative force
A potential energy function can be defined for any conservative force, including Coulomb force
The notions of potential and potential energy are important for practical problem solving
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The electrostatic force is conservative
As in mechanics, work is
Work done on the positive charge by moving it from A to B
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A B
E
d
cosW Fd
cosW Fd qEd
The work done by a conservative force equals the negative of the change in potential energy, DPE
This equation is valid only for the case of a uniform electric field
allows to introduce the concept of electric potential
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PE W qEdD
The potential difference between points A and B, VB-VA, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge
Electric potential is a scalar quantity Electric potential difference is a measure of electric
energy per unit charge Potential is often referred to as “voltage”
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B A
PEV V V
q
DD
Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential
In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V
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1 1V J C
Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter)
Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus, A positive charge gains electric potential energy when it is
moved in a direction opposite the electric field
A negative charge looses electrical potential energy when it moves in the direction opposite the electric field
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1 1N C V m
The same kinetic-potential energy theorem works here
If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energy
If a negative charge is released from A, it accelerates in the direction opposite the electric field
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A
B
q d
A
B
m d
E g
i i f fKE PE KE PE
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Vab
What is the speed of an electron accelerated from rest across a
potential difference of 100V? What is the speed of a proton
accelerated under the same conditions?
Given:
DV=100 V
me = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Find:
ve=?
vp=?
Observations:
1. given potential energy
difference, one can find the
kinetic energy difference
2. kinetic energy is related to
speed
i i f fKE PE KE PE
f i fKE KE KE PE q V D D
21 2
2f f
q Vmv q V v
m
D D
6 55.9 10 , 1.3 10e pm mv v
s s
Electric circuits: point of zero potential is defined by grounding some point in the circuit
Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge
With this choice, a potential can be found as
Note: the potential depends only on charge of an object, q, and a distance from this object to a point in space, r.
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e
qV k
r
If more than one point charge is present, their electric potential can be found by applying superposition principle
The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges.
Remember that potentials are scalar quantities!
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Consider a system of two particles
If V1 is the electric potential due to charge q1 at a point P, then work required to bring the charge q2 from infinity to P without acceleration is q2V1. If a distance between P and q1 is r, then by definition
Potential energy is positive if charges are of the same sign and vice versa.
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P A
q1 q2
r
1 22 1 e
q qPE q V k
r
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Three ions, Na+, Na+, and Cl-, located such, that they
form corners of an equilateral triangle of side 2 nm in
water. What is the electric potential energy of one of the
Na+ ions?
Cl-
Na+ Na+
? Na Cl Na Na Na
e e e Cl Na
q q q q qPE k k k q q
r r r
but : !Cl Naq q
0Nae Na Na
qPE k q q
r
Recall that work is opposite of the change in potential energy,
No work is required to move a charge between two points that are at the same potential. That is, W=0 if VB=VA
Recall: 1. all charge of the charged conductor is located on its surface 2. electric field, E, is always perpendicular to its surface, i.e. no
work is done if charges are moved along the surface
Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium
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B AW PE q V V
… but that’s not all!
Because the electric field in zero inside the conductor, no work is required to move charges between any two points, i.e.
If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor
Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface!
Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero!
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0B AW q V V
A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (eV)
The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V
Relation to SI:
1 eV = 1.6010-19 C·V = 1.6010-19 J
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Vab=1 V
Remember that potential is a scalar quantity Superposition principle is an algebraic sum of potentials due
to a system of charges
Signs are important
Just in mechanics, only changes in electric potential are significant, hence, the point you choose for zero electric potential is arbitrary.
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In the Bohr model of a hydrogen atom, the electron, if it is in the
ground state, orbits the proton at a distance of r = 5.2910-11 m. Find
the ionization energy of the atom, i.e. the energy required to remove
the electron from the atom.
Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not
a very good model of the atom. A better picture is one in which the electron is spread
out around the nucleus in a cloud of varying density; however, the Bohr model does
give the right answer for the ionization energy
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In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,
orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,
i.e. the energy required to remove the electron from the atom.
Given:
r = 5.292 x 10-11 m
me = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Find:
E=?
The ionization energy equals to the total energy of the
electron-proton system,
E PE KE
22 2182.18 10 J -13.6 eV
2 2
ee e
k ee m eE k k
r mr r
The velocity of e can be found by analyzing the force
on the electron. This force is the Coulomb force;
because the electron travels in a circular orbit, the
acceleration will be the centripetal acceleration:
c cma F
2 2
,2
e
e vPE k KE m
r with
or
2 2
2,e
v em k
r r or
22 ,e
ev k
mr
Thus, total energy is
They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage)
The electric field at every point of an equipotential surface is perpendicular to the surface
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convenient to represent by drawing
equipotential lines
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ˆˆ ˆ( )dV V V V
E i j kdr x y z
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23.11 .. Three point charges, which initially are infinitely far apart, are placed at the
corners of an equilateral triangle with sides d.Two of the point charges are identical and
have charge q. If zero net work is required to place the three charges at the corners of
the triangle, what must the value of the third charge be?
23.27 .. A thin spherical shell with radius R1 = 3.00cm is concentric with a larger
thin spherical shell with radius R2 = 5.00cm. Both shells are made of insulating
material. The smaller shell has charge q1 = +6.00nC distributed uniformly over its
surface, and the larger shell has charge q2 = -9.00nC distributed uniformly over its
surface. Take the electric potential to be zero at an infinite distance from both shells.
(a) What is the electric potential due to the two shells at the following distance from
their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00cm? (b) What is the
magnitude of the potential difference between the surfaces of the two shells? Which
shell is at higher potential: the inner shell or the outer shell?
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1. a) E = 0 V/m in throughout some region of space, can you conclude that
the potential V = 0 in this region?
b) V = 0 V throughout some region of space. Can you conclude that the
electric field E = 0 V/m in this region?
--Find the electric potential everywhere for a sphere (radius
R) with charge (Q) uniformly distributed. Take V=0 at infinity.
--Sketch V vs r and Er vs r.
E =
KQ
R3r ˆ r
KQ
r2ˆ r
r < R
R < r
ì
í ï
î ï
Given
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Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly
distributed.
V =
3KQ
2R-
KQ
2R3r2
KQ
r
(r < R)
(R < r)
ì
í
ï ï
î
ï ï
E r =K
Q
R3r
KQ
r2
(r < R)
(R < r)
ì
í ï
î ï
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Find the x,y and z components of the electric field, given that the electric potential of a
disk is given by
Vdisk =Q
2pR2e0
z2 + R2 - z( )
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Find the z component of the electric field, given that the electric potential of a disk is
given by
Vdisk =Q
2pR2e0
z2 + R2 - z( )
Edisk( )z
= -dV
dz=
h
2e0
1-z
z2 + R2
æ
èç
ö
ø÷
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is perp to equipotential surfaces
points downhill (decreasing V)
--strength proportional to spacing equipotentials
Geometry of potential/field
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--field is zero inside conductor
--field is perpendicular at surface
--conductor is at equipotential (no work to move)
Conductor in equilibrium: field and potential
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--equipotentials are parallel
to nearby conductor
Conductor in equilibrium: equipotentials
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--Find the electric potential everywhere for a point charge (q) at the center of a hollow
metal sphere (inner radius a, outer radius b) with charge Q. (Take V = 0 at infinity.)
--Sketch V vs r and Er vs r.
a
b
Problem: Finding Potential
E r(r) =
Kq
r2, r < a
0, a < r < b
K(Q + q)
r2, b < r
ì
í
ï ï
î
ï ï
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Problem: Finding Potential (ans)
E r(r) =
Kq
r2, r < a
0, a < r < b
K(Q + q)
r2, b < r
ì
í
ï ï
î
ï ï
r
d
r Vr V∞
∆V
Vr
Va
∆V
E
For all r
For b < r
= - -K(Q + q)
r
é
ë ê ù
û ú ¥
r
=K(Q + q)
r-
K(Q + q)
¥
Defined V¥ = 0
Vr - V¥ =K(Q + q)
r
Vr =K(Q + q)
r(R < r)
Finding Vr
For a < r < b
from b < r; Vb =K(Q + q)
b
Vr =K(Q + q)
b(a < r < b)
For r < a
= - -Kq
r
é
ë ê ù
û ú a
r
=Kq
r-
Kq
a
From before Va = Vb =K(Q + q)
b
Vr -K(Q + q)
b=
Kq
r-
Kq
a
Vr =K(Q + q)
b+
Kq
r-
Kq
a
æ
è
ö
ø (r < a)
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V =
Kq + Q
r
Kq + Q
b
Kq + Q
b+ Kq
1
r-
1
a
æ
è
ö
ø
ì
í
ï ï ï
î
ï ï ï
b < r
a < r < b
r < a
not origin
Problem: finding
Potential (Answer)
E r(r) =
Kq
r2, r < a
0, a < r < b
K(Q + q)
r2, b < r
ì
í
ï ï
î
ï ï
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-charge separation
-not sustained
Sources of potential: Capacitor
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definition
E =
KQ
R3r ˆ r
KQ
r2ˆ r
r < R
R < r
ì
í ï
î ï
For all r
For R < r
= - -KQ
r
é
ë ê ù
û ú ¥
r
=KQ
r-
KQ
¥
Defined V¥ = 0
Vr - V¥ =KQ
r
Vr =KQ
r(R < r)
Finding Vr
For r < R
= -KQ
2R3r2é
ë ê ù
û ú R
r
= -KQ
2R3r2 -
KQ
2R3R2æ
è
ö
ø
From R < r : VR =KQ
R=
KQ
R3R2
Vr -KQ
R= -
KQ
2R3r2 -
KQ
2R3R2æ
è
ö
ø
Vr =3KQ
2R3R2 -
KQ
2R3r2 (r < R)
r
∆V
Vr
Vr
E
d
r Vr V∞
VR
∆V