1. General Physics (PHY 2140) Lecture 2 Electrostatics
Electric flux and Gauss’s law
Electrical energy
potential difference and electric potential
potential energy of charged conductors
Chapters 15-16
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2. Lightning Review Last lecture: Coulomb’s law
the superposition principle
The electric field
Review Problem: A “free” electron and “free” proton are placed in an identical electric field. Compare the electric force on each particle. Compare their accelerations.
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3. Review Solution: Recall
For a proton or an electron the size of the force is the same!
Charges are the same in magnitude.
Opposite in sign.
The direction of the electric forces are opposite.
However the accelerations are different!
Mass of electron is 9.11x10-31 Kg
Mass of a proton is 1.67x10-27 Kg
So, the acceleration of the proton is smaller by me/mp = 5.5x10-4
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4. 15.5 Electric Field Lines
A convenient way to visualize field patterns is to draw lines in the direction of the electric field.
Such lines are called field lines.
Remarks:
Electric field vector, E, is tangent to the electric field lines at each point in space.
The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region.
E is large when the field lines are close together and small when far apart.
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5. 15.5 Electric Field Lines (2)
Electric field lines of single positive (a) and (b) negative charges. a) b) + q - q
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6. 15.5 Electric Field Lines (3)
Rules for drawing electric field lines for any charge distribution.
Lines must begin on positive charges (or at infinity) and must terminate on negative charges or in the case of excess charge at infinity.
The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge.
No two field lines can cross each other.
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7. 15.5 Electric Field Lines (4)
Electric field lines of a dipole. + -
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8. 15.6 Conductors in Electrostatic Equilibrium
Good conductors (e.g. copper, gold) contain charges (electron) that are not bound to a particular atom, and are free to move within the material.
When no net motion of these electrons occur the conductor is said to be in electro-static equilibrium.
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9. 15.6 Conductors in Electrostatic Equilibrium
Properties of an isolated conductor (insulated from the ground).
Electric field is zero everywhere within the conductor.
Any excess charge on an isolated conductor resides entirely on its surface.
The electric field just outside a charged conductor is perpendicular to the conductor’s surface.
On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest – at sharp points.
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10. Faraday’s ice-pail experiment+ + - + +
Demonstrates that the charge resides on the surface of a conductor.
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11. Mini-quiz
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? ? + q +
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12. Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge?
Answer:
Negative charge will build up on the inside of the shell.
Positive charge will build up on the outside of the shell.
There will be no field lines inside the conductor but the field lines will remain outside the shell. + + + - - - + + - - + - + + q - + - - + + - - - + + +
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13. 15.9 The oscilloscope
Changing E field applied on the deflection plate (electrodes) moves the electron beam. V2 V1  d L
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14. Oscilloscope: deflection angle (additional)V2 V1  d L
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15. 15.9 Electric Flux and Gauss’s Law
A convenient technique was introduced by Karl F. Gauss ( ) to calculate electric fields.
Requires symmetric charge distributions.
Technique based on the notion of electrical flux.
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16. 15.9 Electric Flux
To introduce the notion of flux, consider a situation where the electric field is uniform in magnitude and direction.
Consider also that the field lines cross a surface of area A which is perpendicular to the field.
The number of field lines per unit of area is constant.
The flux, F, is defined as the product of the field magnitude by the area crossed by the field lines.
Area=A E
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17. 15.9 Electric Flux Units: Nm2/C in SI units.
Find the electric flux through the area A = 2 m2, which is perpendicular to an electric field E=22 N/C
Answer: F = 44 Nm2/C.
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18. 15.9 Electric Flux
If the surface is not perpendicular to the field, the expression of the field becomes:
Where q is the angle between the field and a normal to the surface. N q q
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19. 15.9 Electric Flux
Remark:
When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive.
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20. Example:
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”. y 1 2 E x z
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21. Dealing with a composite, closed surface.
Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
Dealing with a composite, closed surface.
Sum of the fluxes through all surfaces.
Flux of field going in is negative
Flux of field going out is positive.
E is parallel to all surfaces except surfaces labeled 1 and 2.
So only those surfaces (1 & 2) contribute to the flux. x y z E 1 2
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22. Solution: y 1 2 E x z Question:
Calculate the flux of a constant E field (along x) through a cube of side “L”.
Reasoning:
Dealing with a composite, closed surface.
Sum of the fluxes through all surfaces.
Flux of field going in is negative
Flux of field going out is positive.
E is parallel to all surfaces except surfaces labeled 1 and 2.
So only those surface contribute to the flux.
Solution: x y z E 1 2
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23. 15.9 Gauss’s Law
The net flux passing through a closed surface surrounding a charge Q is proportional to the magnitude of Q:
In free space, the constant of proportionality is 1/eo where eo is called the permittivity of of free space.
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24. 15.9 Gauss’s Law
The net flux passing through any closed surface is equal to the net charge inside the surface divided by eo.
Can be used to compute electric fields. Example: point charge
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25. 16.0 Introduction 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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26. 16.1 Potential difference and electric potential
The electrostatic force is conservative
As in mechanics, work is
Work done on the positive charge by moving it from A to B A B d
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27. Potential energy of electrostatic field
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 us to introduce the concept of the electric potential
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28. Electric potential
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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29. Electric potential - units
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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30. Electric potential - notes
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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31. Analogy between electric and gravitational fields
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 A A d d q m B B
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32. Example: motion of an electron
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?
Observations:
given potential energy difference, one can find the kinetic energy difference
kinetic energy is related to speed
Given:
DV=100 V
me = 9.11´10-31 kg
mp = 1.67´10-27 kg
|e| = 1.60´10-19 C
Find:
ve=?
vp=?
Vab
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33. 16.2 Electric potential and potential energy due to point charges
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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34. Superposition principle for potentials
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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35. Potential energy of a system of point charges
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. q2 q1 r P A
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36. Mini-quiz: potential energy of an ion
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+
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37. Recall from last chapter:
Electric field lines of a dipole. + -
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38. 16.3 Potentials and charged conductors
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:
all charge of the charged conductor is located on its surface
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
… but that’s not all!
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39. 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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40. The electron volt
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 = 1.60´10-19 C·V = 1.60´10-19 J
Vab=1 V
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41. Problem-solving strategy
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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42. Example : ionization energy of the electron in a hydrogen atom
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´10-11 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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43. 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 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom.
Given:
r = x m
me = 9.11´10-31 kg
mp = 1.67´10-27 kg
|e| = 1.60´10-19 C
Find:
E=?
The ionization energy equals to the total energy of the electron-proton system,
with
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: or or
Thus, total energy is
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44. 16.4 Equipotential surfaces
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
convenient to represent by drawing equipotential lines
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