1. Chapter 25 Electric Potential
25.1 Potential Difference and Electric Potential
25.2 Potential Differences in a Uniform Electric Field
25.3 Electric Potential and Potential Energy Due to Point charges
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2. 25.1 Potential Difference and Electric Potential
When a test charge q0 is placed in an electric field E created by some source charge distribution, the electric force acting on the test charge is q0 E.
When the test charge is moved in the field by some external agent, the work
done by the field on the charge is equal to the negative of the work done by the external agent causing the displacement
This is analogous to the situation of lifting an object with mass in a gravitational field—the work done by the external agent is mgh and the work done by the gravitational force is -mgh.
the work done by the electric field on the charge is
work =F. ds =q0E“.ds
ds is the displacement of a charge
The potential energy of the charge–field system is changed by an amount dU = q0E“.ds
For a finite displacement of the charge from point A to point B, the change in potential energy of the system ∆U =UB - UA is
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3. field at constant velocity is
Dividing the potential energy by the test charge gives a physical quantity that depends only on the source charge distribution. The potential energy per unit charge U/q0 is
independent of the value of q0 and has a value at every point in an electric field. This quantity U/q0 is called the electric potential (or simply the potential) V.
The potential difference ∆V =VB -VA between two points A and B in an electric field is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge q0:
The potential difference between A and B depends only on the source charge distribution (consider points A and B without the presence of the test charge)
the work done by an external agent in moving a charge q through an electric
field at constant velocity is
W= q ∆V
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4. Definition of the Electric Potential
The electric potential energy of a charged particle in an electric field depends not only on the electric field but on the charge of the particle.
We want to define a quantity to probe the electric field that is independent of the charge of the probe.
We define the electric potential as
Unlike the electric field, which is a vector, the electric potential is a scalar.
The electric potential has a value everywhere in space but has no direction.
“potential energy per unit charge of a test particle”
Units: [V] = J / C, by definition, volt
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5. Electric Potential Energy
The electric force, like the gravitational force, is a conservative force.
When an electrostatic force acts between two or more charges within a system, we can define an electric potential energy, U, in terms of the work done by the electric field, We, when the system changes its configuration from some initial configuration to some final configuration.
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6. The negative sign on the work: If E does positive work then U < 0
Like gravitational or mechanical potential energy, we must define a reference point from which to define the electric potential energy.
We define the electric potential energy to be zero when all charges are infinitely far apart.
We can then write a simpler definition of the electric potential taking the initial potential energy to be zero,
The negative sign on the work:
If E does positive work then U < 0
If E does negative work then U > 0
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7. Constant Electric Field - Special Cases
Displacement is in the same direction as the electric field
A positive charge loses potential energy
when it moves in the direction of the electric field.
Displacement is in the direction opposite to the electric field
A positive charge gains potential energy when it moves in the direction opposite to the electric field.
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8. Potential Difference and Electric Potential
The potential difference between points A and B, VB – VA, is defined as the change in potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge.
DV VB – VA = DU q
Final point
Initial point
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10. Work WAB = -DU = - q(VB – VA) Initial point Final point Final point
Equation is true if the only force is the conservative electrostatic force. That is, there are no non conservative forces acting on the system. 10
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11. Change in Potential Energy
As the electric field accelerates the charge, the charge gains kinetic energy.
As the charged particle gains kinetic energy, it loses an equal amount of potential energy DK = - DU
By definition, the work done by a conservative force equals the negative change in potential energy, DU DU = - WAB = - qEd
This equation is valid only for a uniform electric field.
Potential Energy
DU = Ub – Ua = qVba
Final point
Initial point 11
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12. The definition of work is
25-2 Potential Differences in a Uniform Electric Field (Constant Electric Field )
Let’s look at the electric potential energy when we move a charge q by a distance d in a constant electric field.
The definition of work is
For a constant electric field the force is F = qE
the work done by the electric field on the charge is
q = angle between E and d.
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13. Work and Potential Energy
There is a uniform field between the two plates
As the positive charge moves from A to B, work is done
WAB=F d=q E d
ΔPE =-W AB=-q E d
only for a uniform field
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14. Important! Usually take “low” side as V = 0.
The positive side is always the “high” potential side, regardless of the sign of the charge.
The definition of the “high” side is done for a positive test charge.
Important!
Usually take “low” side as V = 0.
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15. Conservative Forces
A force is conservative if the work it does on a particle moving between any two points is independent of the path taken by the particle.
The work done by a conservative force exerted on a particle moving through any closed path is zero.
Gravitational (Newton’s law of gravity) and Electrical (Coulomb’s law of electrical force) are both conservative forces.
Since electrostatic force is conservative, electrostatic phenomena can be described in terms of electrical potential energy.
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16. Potential Diff. In Uniform E field
Charged particle moves from A to B in uniform E field.
DV =
E . ds = =  E . d B A q
DU = qo DV = - qo E . d =  qo E d cos q
Show that the potential diff. between path (1) and (2) are the same as expected for a conservative force field.
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17. Potential Diff. In Uniform E field (Path independence)
Show that the potential difference between path (1) and (2) are the same as expected for a conservative force field. q d
path (1) B A
DV = -
E . ds =  Es cos q
path (1)
path (2)
= since E ds B c c A
path (2)
DV = -
E . ds E . ds
Same c A
DV = 
E . ds =  Ed=  E s cos q
Conservative force: The work is path-independent.
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18. Electric Potential The Electron Volt 1 V = 1 J/C
Unlike the electric field, which is a vector, the electric potential is a scalar.
The electric potential has a value everywhere in space but has no direction.
The SI units of electric potential are joules per coulomb.
The unit of potential is the volt.
A VOLT is defined:
Units: [V] = J / C, by definition, volt
1 V = 1 J/C
The Electron Volt
The electron volt is defined as the energy acquired by a particle carrying a charge equal to that on the electron (q = e) as the result of moving through a potential difference of 1 V.
1eV = (1.6 x C)(1.0 V) = 1.6 x J
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19. Equipotential
DV = 
E . ds = 0 B c
VC = VB ( same potential)
In fact, points along this line has the same potential. We have an equipotential line.
If s is perpendicular to E (path C-B), the electric potential does not change.
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20. Equipotential Surfaces
The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.
No work is done in moving a test charge between any two points on an equipotential surface.
The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.
Equipotential surfaces are always perpendicular to electric field lines.
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21. Equipotential Surface
Equipotential Surfaces (dashed blue lines) and electric field lines (orange lines) for (a) a uniform electric field produced by infinite sheet of charge, (b) a point charge, and (c) an electric dipole. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.
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22. Equipotential Surfaces
The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.
No work is done in moving a test charge between any two points on an equipotential surface.
The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.
Equipotential surfaces are always perpendicular to electric field lines.
Lines of constant E
Lines of constant V
(perpendicular to E) 22
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23. Equipotential Surfaces and the Electric Field
All points on the surface of a charged conductor in electrostatic equilibrium are at the same potential
Therefore, the electric potential is a constant everywhere on the surface of a charged conductor in equilibrium
An ideal conductor is an equipotential surface. Therefore, if two conductors are at the same potential, the one that is more curved will have a larger electric field around it. This is also true for different parts of the same conductor.
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24. Example
A point particle of mass m = 1.8x10-5 kg and charge q = +3.0x10-5 C is released from rest at point A and accelerates until it reaches point B. The only force acting on the particle is the electric force and the electric potential at A is 25 V greater than at B. What is the speed of the particle when it reaches B?
solve for vB
What happens if q is negative?
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25. Example
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 10 –17 J. Calculate the charge on the ion.
qV= 7.37x10-17 J , V=115 V
q = 6.41x10-19 C
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26. Example
In a hydrogen atom the e- revolves around the p+ at a distance of 5.3 x m. Find the electric potential at the e- due to the p+, and the electrostatic potential energy between them.
Electric potential due to proton:
Electrostatic p.E. is given by: p+
e - r
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27. Example
A proton is released from rest in a uniform E field that has a magnitude of 8 x 104 V/m and is directed along the positive x-axis. The proton undergoes a displacement of 0.50 m in the direction of E.
Find the change in electric potential between points A and B.
Find the change in potential energy of the proton for this displacement.
DV = Ed =  (8.0x104 V/m) (0.50m) =  4.0x104 V
DU = q DV = (1.6 x C) (4 .0x104 V) = 6.4 x J
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28. Example
Suppose an electron is released from rest in a uniform electric field whose magnitude is 5.90 x 103 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm?
(a) |DV| = Ed = (5.90 x 103 V/m)( m) = V
(b) q |DV| = mv2/2  v = 4.55x106 m/s
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29. Example
Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x m.
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30. Forces on Charged Particles
Example
Forces on Charged Particles
In a CRT an electron moves 0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find:
(a) The force on the electron.
(b) The work done on it by the E-field.
(c) Its potential difference from start to finish.
(d) Its change in potential energy.
(e) Its final speed.
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31. Example (a) Force is in opposite direction to the E-field, magnitude:
(b) Work done by force:
(c) Potential difference is defined as work/unit charge:
Alternatively (e- opposite to p+):
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32. (d) Change in potential energy:
(e) Loss of PE = gain in KE = ½mv2
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33. Example
A proton is accelerated across a potential difference of 600 V. Find its change in K.E. and its final velocity.
By definition, 1 eV = 1.6 x J.
Acceleration across 600 V
Proton gains 600 eV.
DK.E. = 600(1.6 x 10-19) = 9.6 x J
Final velocity is:
If it started from rest
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34. Example
Two parallel metal plates have an area A = 225 cm2 and are d =0.5 cm apart, with a p.d. of 0.25 V between them. Calculate the electric field. 0V
0.25V
0.1V
0.2V
x =0
x =0.5m ds
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35. Electric Potential and Potential Energy due to point charges
Consider isolated positive point charge q (i.e. E directed radially outward from the charge)
To find electric potential at a point located at a distance r from the charge, start with the general expression for potential difference:
VB VA =
E . ds B A
Where A and B are two arbitrary points as shown.
E = kq/r2 r, where r is a unit vector directed from the charge toward the field point.
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36. Electric Potential and Potential Energy due to point charges
We can express E . ds as
E . ds = kq/r2 r . ds
The magnitude of r is 1, dot product r . ds = ds cos q, where q is the angle between r and ds .
ds cos q is the projection of ds onto r , thus
ds cos q = dr. B A B A
VB-VA = -
E . ds
= -
kq/r2 dr
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37. Electric Potential and Potential Energy due to point chargesrB rA B A
VB-VA = -
Er dr
= -
kq/r2 dr rB kq
VB-VA = r rA 1 1
VB-VA = kq rB rA
Depends only on the coordinates and not on the path.
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38. Electric Potential and Potential Energy due to point charges
rA = infinity (and VA = 0), we have electric potential created by a point charge at a distance r from the charge given by
V = kq r
Points at same distance r from q have the same potential V, i.e. the equipotential surfaces are spherical and centered on the charge.
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39. Potential due to two or more charges: Superpositionq1 r1 q5 q2
V = qi ri S k i r5 r2 P r3 r4 q3 q4
where potential is taken to be zero at infinity and ri is the distance from the point P to the charge qi.
Note that this is a scalar sum rather than a vector sum.
The potential is positive if the charge is positive and negative if the charge is negative
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40. Example Finding the Electric Potential at Point P
Superposition: Vp=V1+V2
Vp=1.12104 V+(-3.60103 V)=7.6103 V
5.0 mC
-2.0 mC
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41. Example:
How many electrons should be removed from an initially uncharged spherical conductor of radius m to produce a potential of 7.5 kV at the surface? =
= 7.50 x 103 V
N = 1.56 x electrons
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42. Example
A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x axis. For what finite value(s) of x is (a) the electric field zero ? (b) the electric potential zero ?
x x – 4.00 = 0
(x+4.83)(x0.83)=0
x = m
(other root is not physically valid)
x = m and x= m
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43. Potential Energy of a system of two charges
V1 = potential at a point P due to q1, external agent must do work to bring a second charge q2 from infinity to P and this work = q2V1.
Definition: This work done is equal to the potential energy U of the two-particle system.
If two point charges are separated by a distance r12, the potential energy of the pair of charges is given by
U12 = k q1 q2
r12
k = 1
4peo
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44. Potential Energy
Three point charges are fixed at the positions shown. The potential energy of this system of charges is given by
U = k q1 q2
r12 q3
r13
r23 +
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45. If the charges have the same sign, PE is positive
Notes About Electric Potential Energy of Two Charges
If the charges have the same sign, PE is positive
Positive work must be done to force the two charges near one another
The like charges would repel
If the charges have opposite signs, PE is negative
The force would be attractive
Work must be done to hold back the unlike charges from accelerating as they are brought close together
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46. Example
What is the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s?
U = 0 + U12 + (U13 + U23) + (U14 + U24 + U34) s s s s
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47. Example
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.
E = 105 V/m
d = 10 cm
a b +
V = Vb – Va = -Ed
U = q V
K = (1/2)mv2
U + K = 0
K = -U
(1/2)mv2 = -q V = +qEd
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48. (a) The force on the electron. (b) The work done on it by the E-field.
Example
In a CRT an electron moves 0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find:
(a) The force on the electron.
(b) The work done on it by the E-field.
(c) Its potential difference from start to finish.
(d) Its change in potential energy.
(e) Its final speed.
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49. (a) Force is in opposite direction to the E-field, magnitude:
(b) Work done by force:
(c) Potential difference is defined as work/unit charge:
Alternatively (e- opposite to p+):
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50. (d) Change in potential energy:
(e) Loss of PE = gain in KE = ½mv2
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51. Example
Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x m.
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52. Example
The charge distribution as shown is referred to as a linear quadrupole. (a) What is the potential at a point on the axis where x > a? (b) What happens when x >> a?
V = ke Q =
V =
As x >> a
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53. Example
When a negative charge moves in the direction of the electric field,
1. the field does positive work on it and the potential energy increases
2. the field does positive work on it and the potential energy decreases
3. the field does negative work on it and the potential energy increases
4. the field does negative work on it and the potential energy decreases
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54. Example
The electric potential energy of two point charges approaches zero as the two point charges move farther away from each other.
If the three point charges shown here lie at the vertices of an equilateral triangle, the electric potential energy of the system of three charges is
1. positive
2. negative
3. zero
4. not enough information given to decide
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55. Example
The electric potential due to a point charge approaches zero as you move farther away from the charge.
If the three point charges shown here lie at the vertices of an equilateral triangle, the electric potential at the center of the triangle is
1. positive
2. negative
3. zero
4. not enough information given to decide
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56. E-field between two parallel plates
Assume uniform field, and 3 mm plate separation
E = |VB – VA| / d = 12 / 3.0x10-3 = 4000 V/m
E directed from A (+ve) to B (ve)
A(+ve) plate is at higher potential than –ve plate.
Potential difference between plates = potential difference between battery terminals because all points on a conductor in equilibrium are at the same electric potential; no potential difference exists between a terminal and any portion of the plate to which it is connected.
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57. Example
If a positive charge be moved against the electric field, then what will happen to the energy of the system?
If a positive charge be moved against the electric field, then energy will be used from an outside source.
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58. Example
If 80 J of work is required to transfer 4 C charge from infinity to a point, find the potential at that point
W =80 J, q = 4 C, V =?
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59. Electric Field and Electric Potential
Example
Electric Field and Electric Potential
4. Which of the following figures have V=0 and E=0 at red point? q q -q A B q q -q q -q C D E
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60. Example What is the electric potential at a distance of 0.529 A from
the proton?
V = kq/r = 9x109 N m2//C2 x1.6x10-19 C/0.529 x10-10m
V = J/C = Volts
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61. Example
Two test charges are brought separately to the vicinity of a positive charge Q A q r Q
Charge +q is brought to pt A, a distance r from Q B Q 2q 2r
Charge +2q is brought to pt B, a distance 2r from Q
(a) UA < UB
(b) UA = UB
(c) UA > UB
I) Compare the potential energy of q (UA) to that of 2q (UB)
The potential energy of q is proportional to Qq/r
The potential energy of 2q is proportional to Q(2q)/(2r) = Qq/r
Therefore, the potential energies UA and UB are EQUAL!!!
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62. Example
(a)
(b)
(c)
II) Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞ ?
The principle at work here is CONSERVATION OF ENERGY.
Initially:
The charge has no kinetic energy since it is at rest.
The charge does have potential energy (electric) = UB.
Finally:
The charge has no potential energy (U µ 1/R)
The charge does have kinetic energy = KE
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63. Example
: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of .0529nm, and the next level has a radius of .212m.
What is the potential difference between the two levels?
Which level has a higher potential? +e r1 r2
r1 is at a higher potential.
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64. Example
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65. Example What is the electric potential difference between A and B?
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66. Example Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.
what is the electric potential at points A and B?
The potential is a scalar (not vector) sum of the electric potentials produced by the individual charges:
Electric Potential at A:
Electric Potential at B:
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67. Example Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.
(b) what is the work done by the electric field on a point charge of 2.50 nC that travels from A to B?
V = EPE/q0
WAB = EPEA – EPEB
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68. Example
What is the total energy of the electron in a hydrogen atom? In the ground state the electron orbit about the proton has a radius equal to one Bohr radius rB = 5.29x10-11 m.
qe = -qp = -1.60x10-19 C, me = 9.11x10-31 kg and ve = 2.2x106 m/s
But the electron in a hydrogen atom also has kinetic energy:
Total energy of e- in hydrogen atom is EPE + KE = -2.15x10-18 J
This is the electron’s binding energy, i.e., how much energy is required to rip off an electron!
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69. Example:
Three identical point charges (q = +2.0 μC each) are brought from infinity and fixed to a straight line so that the spacing between adjacent charges is d = 0.40 m. Determine the electric potential energy of this group. q q q
U = k q1 q2
r12 q3
r13
r23 + d d
U = k
0.4
0.8 +
2x10-6
x2x10-6
2x10-6
x2x10-6
2x10-6
x2x10-6
EPE = 0.23 J
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70. Example
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71. Example Points A, B, and C lie in a uniform electric field.
What is the potential difference between points A and B?
ΔVAB = VB - VA
a) ΔVAB > 0 b) ΔVAB = 0 c) ΔVAB < 0
The electric field, E, points in the direction of decreasing potential
Since points A and B are in the same relative horizontal location in the electric field there is on potential difference between them
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72. Example Points A, B, and C lie in a uniform electric field.
Point C is at a higher potential than point A.
True False
As stated previously the electric field points in the direction of decreasing potential
Since point C is further to the right in the electric field and the electric field is pointing to the right, point C is at a lower potential
The statement is therefore false
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73. Therefore the electric potential energy also doesn’t change
Example E A B C
Points A, B, and C lie in a uniform electric field.
If a negative charge is moved from point A to point B, its electric potential energy
a) Increases. b) decreases. c) doesn’t change.
The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location
As shown in Example, the potential at points A and B are the same
Therefore the electric potential energy also doesn’t change
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74. Example Points A, B, and C lie in a uniform electric field.
Compare the potential differences between points A and C and points B and C.
a) VAC > VBC b) VAC = VBC c) VAC < VBC
In Example 4 we showed that the the potential at points A and B were the same
Therefore the potential difference between A and C and the potential difference between points B and C are the same
Also remember that potential and potential energy are scalars and directions do not come into play
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75. Example
A positive charge is released from rest in a region of electric field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
A positive charge placed in an electric field will experience a force given by
But E is also given by
Therefore
Since q is positive, the force F points in the direction opposite to increasing potential or in the direction of decreasing potential
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76. Example
If you want to move in a region of electric field without changing your electric potential energy. You would move
Parallel to the electric field
Perpendicular to the electric field
The work done by the electric field when a charge moves from one point to another is given by
The way no work is done by the electric field is if the integration path is perpendicular to the electric field giving a zero for the dot product
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77. Example
Electric force moves a charge of x10-10 C from point A to point B and does 5x10-6 J of work.
What is the difference in potential energies of A and B (EPEA – EPEB)?
EPEA – EPEB = 5x10-6 J
What is the potential difference between A and B (VA – VB)?
V = V Point A is higher potential
WAB = - (EPEB - EPEA )
∆PE: = EPEA - EPEB = 5x10-6 J
∆V: VA – VB = (EPEA – EPEB ) / q= 5x10-6J/2x10-10C = V
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78. Example What is the electric potential at the center of the square? r
45º r
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79. A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C.
How much work would be required to move a proton from the negative to the positive plate?
What is the potential difference between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
Example 1
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80. b) What is the potential difference between the plates?
Example
A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C.
b) What is the potential difference between the plates?
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81. Example
A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C.
c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
Norah Ali Almoneef

82. Example
Calculate the electric potential energy between each pair of charges and add them together.
Norah Ali Almoneef

83. None of these is correct.
EXAMPLE
Charges +Q and –Q are arranged at the corners of a square as shown. When the magnitude of the electric field E and the electric potential V are determined at P, the center of the square, we find that
E ≠ 0 and V > 0.
E = 0 and V = 0.
E = 0 and V > 0.
E ≠ 0 and V < 0.
None of these is correct.
Norah Ali Almoneef

84. EXAMPLE
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The potential for line c is
100 V.
100 V.
200 V.
200 V.
zero
Norah Ali Almoneef

85. 100 eV. 100 eV. 200 eV. 200 eV. zero EXAMPLE
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The work required to move a third charge, q = e, from the 100 V line to b is
100 eV.
100 eV.
200 eV.
200 eV.
zero
Norah Ali Almoneef

86. The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point is tripled, the potential becomes
EXAMPLE
V/3.
3V.
V/9.
9V.
1/V 2 .
Norah Ali Almoneef

87. EXAMPLE
A proton is released from rest in a uniform electric field that has a magnitude of V/m and is directed along the positive x axis. The proton undergoes a displacement of 0.50 m in the direction of E.
(a) Find the change in electric potential between points A and B.
(b) Find the change in potential energy of the proton for
this displacement.
(a)
(b)
The negative sign means the potential energy of the proton decreases as it moves in the direction of the electric field. As the proton accelerates in the direction of the field, it gains kinetic energy and at the same time loses electric potential energy (because energy is conserved).
Norah Ali Almoneef

88. EXAMPLE
A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton
1. moves toward A with an increasing speed.
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
Norah Ali Almoneef

89. EXAMPLE
Norah Ali Almoneef

90. EXAMPLE
Norah Ali Almoneef

91. Summary Potential Difference
The potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.
Potential Difference: VAB = VA - VB
WorkAB = q(VA – VB) Work BY E-field
The positive and negative signs of the charges may be used mathematically to give appropriate signs.
Norah Ali Almoneef

92. Summary Parallel Plates
Consider Two parallel plates of equal and opposite charge, a distance d apart. VA VB E +q
F = qE
Constant E field: F = qE
Work = Fd = (qE)d
Also, Work = q(VA – VB)
So that: qVAB = qEd and
VAB = Ed
The potential difference between two oppositely charged parallel plates is the product of E and d.
Norah Ali Almoneef

93. Summary of Formulas Electric Potential Energy and Potential
Electric Potential Near Multiple charges:
WorkAB = q(VA – VB) Work BY E-field
Oppositely Charged Parallel Plates:
Norah Ali Almoneef

94. Summary Electric potential difference
DV =
E . ds = =  E . d B A
Electric potential difference
Since only differences in potential matter, the location of the zero of the potential can be chosen as we wish
Electric potential of a point charge
If we chose the zero to be infinitely far from a point charge, we can write its potential as
Equipotentials
These are surfaces of equal potential difference. The surface of a conductor in equilibrium is an equipotential.