660 lines
13 KiB
Plaintext
660 lines
13 KiB
Plaintext
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Electricity and Magnetism
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Electric Fields
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Lana Sheridan
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De Anza College
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Jan 12, 2018
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Last time
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• Forces at a fundamental level
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• Electric field
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• net electric field
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• electric field lines
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Warm Up Questions
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Which of the following could be the charge on the particle hidden by the question mark?
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far more work cian and physifields of a hypothetical ssian surface, calculations of ibution. For exsphere with a we discuss in
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dE:
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ace to the
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on a Gaussian ited example, tward from the tely tells us that er a particle or
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Fig. 23-1 A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points, you can conclude that a net positive
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Spherical Gaussian surface
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?
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E
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(A) 0 C (B) −1 C
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(C) −1.6 × 10−19 C (D) +1 μC
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1Figure from Halliday, Resnick, Walker
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Warm Up Questions
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Which of the following could be the charge on the particle hidden by the question mark?
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far more work cian and physifields of a hypothetical ssian surface, calculations of ibution. For exsphere with a we discuss in
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dE:
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ace to the
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on a Gaussian ited example, tward from the tely tells us that er a particle or
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Fig. 23-1 A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points, you can conclude that a net positive
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Spherical Gaussian surface
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?
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E
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(A) 0 C (B) −1 C
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(C) −1.6 × 10−19 C (D) +1 μC ←
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1Figure from Halliday, Resnick, Walker
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Overview
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• Electric field lines
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• Net electric field
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• the effect of fields on charges
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• the electric dipole
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Field Lines
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The electrostatic field caused by an electric dipole system looks something like:
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are intersections of these surfaces with the page) and elecendicular to the electric field lines at every point.
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metric electric a point charge
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An electric field produced by an electric dipole
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c
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Notice that the lines point outward from a positive charge and inward toward a negative charge.
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1Figure from Serway & Jewett
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Field Lines
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es. Figure 23.22 The electric field
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!2q "q
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one that terminates on "q.
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!"
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1Figure from Serway & Jewett
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Field Lines
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Compare the electrostatic fields for two like charges and two opposite charges:
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FIELDS
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tive r.
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e ing age. ec
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ic t it int.
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positive test charge at any point near the of Fig. 22-3a, the net electrostatic force acti the test charge would be perpendicular t sheet, because forces acting in all other tions would cancel one another as a res the symmetry. Moreover, the net force o test charge would point away from the sh shown. Thus, the electric field vector at any in the space on either side of the sheet i perpendicular to the sheet and directed from it (Figs. 22-3b and c). Because the cha uniformly distributed along the sheet, a field vectors have the same magnitude. Such an electric field, with the same nitude and direction at every point, is a uniform electric field. Of course, no real nonconducting sheet (such as a flat expanse of plastic)
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E
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+
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+
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tric field it represents are said to have rotational symmetry about that axis.The electric field vector at one point is shown; note that it is tangent to the field line through that point.
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Fig. 22-5 Field lines for a positive point charge and a nearby negative point charge that are equal in magnitude.The charges at
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fi n
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n n
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s s fi
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2
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T a F
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T
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+
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E
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Field Lines
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Compare the fields for gravity in an Earth-Sun system and electrostatic repulsion of two charges:
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CTRIC FIELDS
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ual positive
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ach other.
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gative
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-dimen
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lly rotate
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xis passing
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of the page.
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d the elec
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avreota
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e electric
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note that it
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that point.
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positive test charg
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of Fig. 22-3a, the ne
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the test charge wo
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sheet, because for
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tions would cance
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the symmetry. Mo
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test charge would
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shown. Thus, the el
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in the space on ei
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perpendicular to
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from it (Figs. 22-3b
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uniformly distribu
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field vectors have the same magnitude. Such an ele
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nitude and direction at every point, is a uniform elec
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E
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+
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+
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14:16 Page 582
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1Gravity figure from http://www.launc.tased.edu.au ; Charge from Halliday, Resnick, Walker
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Field Lines: Uniform Field
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Imagine an infinite sheet of charge. The lines point outward from the positively charged sheet.
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F
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E
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++++
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++++
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++++
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++++
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Positive test charge
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(a) (b)
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+++
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++++
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+++
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+
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++
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+
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+++++++++
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(c)
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++
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+
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-2 were of uniform positive charge, the electric field the sphere would be directed radially away from field lines would also extend radially away from the llowing rule:
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way from positive charge (where they originate) and re they terminate).
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t of an infinitely large, nonconducting sheet (or plane) of positive charge on one side. If we were to place a
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tic force a very unirge on ector e, and ce tend d
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E:
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1Figure from Halliday, Resnick, Walker.
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Electric field due to an Infinite Sheet of Charge
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Suppose the sheet is in air (or vacuum) and the charge density on the sheet is σ (charge per unit area):
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E= σ
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20
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It is uniform! It does not matter how far a point P is from the sheet, the field is the same.
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E
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+
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++
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++
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++
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+++++++++
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+
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)
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Field Lines: Uniform Field
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The field from two infinite charged plates is the sum of each field.
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s
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P0
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(24.8)
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!
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!
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!
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!
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! ! ! ! ! ! ! !
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.14 (Example 24.5) field lines due to an
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!
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!
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!
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!
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! ! ! ! ! ! ! !
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"
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"
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"
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"
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" " " " " " " "
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Figure 24.15 (Example 24.5) The electric field lines between
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The field in the center of a parallel plate capacitor is nearly uniform.
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Free charges in an E-field
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The force on a charged particle is given by F = qE.
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If the charge is free to move, it will accelerate in the direction of the force.
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Example: Ink-jet printing
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Input signals Deflecting plate
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G C Deflecting plate
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E
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e. When a charged oil drop drifted into chamber C through the hole in plate P1, its motion could be controlled by closing and opening switch S and thereby setting up or eliminating an electric field in chamber C. The microscope was used to view the drop, to permit timing of its motion.
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Fig. 22-15 Ink-jet printer. Drops shot
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ing plate P2 chamber C. charged dro ular, our ne By timi and thus de values of q w
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in which e charge, 1.60 quantized, a Modern me experiment
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Ink-Jet Pri
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The need alternative
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Motion of a Charged Particle in an E-field
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n he
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e cd d is a
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(0, 0)
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!
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(x, y)
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vi iˆ !
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! vS
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x
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y
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The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
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S
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ES
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""""""""""""
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!!!!!!!!!!!!
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Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
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Trajectory is a parabola: similar to projectile motion.
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Motion of a Charged Particle in an E-field
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(a) What is the acceleration of an electron in the field of strength E?
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as shown N/C. The
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he elec
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ing one perpen he veloc tric field a curved lectron is tally in a
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ecause the electric field is uniform, a constant electric force is electron, we can model it as a particle under a net force.
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(0, 0)
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!
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(x, y)
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vi iˆ !
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! vS
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x
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y
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The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
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S
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ES
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""""""""""""
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!!!!!!!!!!!!
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Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
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(b) The charge leaves the field at the point (`, yf ). What is yf in terms of `, vi , E , e, and me?
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Motion of a Charged Particle in an E-field
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(a) What is the acceleration of an electron in the field of strength E?
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as shown N/C. The
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he elec
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ing one perpen he veloc tric field a curved lectron is tally in a
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ecause the electric field is uniform, a constant electric force is electron, we can model it as a particle under a net force.
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(0, 0)
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!
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(x, y)
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vi iˆ !
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! vS
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x
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y
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The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
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S
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ES
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""""""""""""
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!!!!!!!!!!!!
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Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
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(b) The charge leaves the field at the point (`, yf ). What is yf in terms of `, vi , E , e, and me?
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yf = − eE `2
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2me v 2
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i
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Trick for working out Net field
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Look for symmetry in the problem.
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To find the E-field, usually several components (Ex , Ey , Ex ) must be found independently.
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If the effects of charges will cancel out, you can neglect those charges.
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If the effects of charges cancel out in one component, just worry about the other component(s).
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Question about net field
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charg magn
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Fig. 22-20 Question 1.
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2 Figure 22-21 shows two square arrays of charged particles. The squares, which are centered on point P, are misaligned. The particles are separated by either d or d/2 along the perimeters of the squares. What are the magnitude and direction of the net electric field at P?
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+6q
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–2q
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+3q –2q
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+3q
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–q
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+6q
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–2q
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–3q
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–q
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+2q –3q
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+2q
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–q
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P
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Fig. 22-21 Question 2.
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5F axis. than a po field their
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1Figure from Halliday, Resnick, Walker, page 597, problem 2.
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Electric Dipole
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electric dipole
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A pair of charges of equal magnitude q but opposite sign, separated by a distance, d.
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dipole moment:
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p = qd rˆ
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where rˆ is a unit vector pointing from the negative charge to the positive charge.
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z
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r(–)
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r(+)
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E(+)
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d
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z
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–q
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+q
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P
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++
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–
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p
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E(–)
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Dipole
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center
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Up here the +q field dominates.
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Down here the –q
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We are usually inte that are large compared that z # d. At such larg proximation, we can neg
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The product qd, wh dipole, is the magnitude of the dipole. (The uni
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The direction of is t dipole, as indicated in
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p:
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p:
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E% q
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4
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&
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'0z
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Electric Dipole (Example 23.6, B)
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Evaluate the electric field from the dipole at point P, which is at
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position (0, y ).
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uq1u 5 uq2u and a 5 b.
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P
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y
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r
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a
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q a –q
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x
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u
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u
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uu
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!"
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ES
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E2
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S
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E1
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S
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ure 23.13 (Example 23.6) en the charges in Figure 2 are of equal magnitude equidistant from the origin, situation becomes symmets shown here.
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y 5 E 1y 1 E 2y 5 ke
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0 q1 0
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a 2 1 y 2 sin f 2 ke
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0 q2 0
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b 2 1 y 2 sin u
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Electric Dipole (Example 23.7)
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P
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y
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r
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a
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q a –q
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x
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u
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u
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uu
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!"
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ES
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E2
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S
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|
E1
|
|||
|
|
S
|
|||
|
|
)
|
|||
|
|
n,
|
|||
|
|
q1 0
|
|||
|
|
1 y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
The y -components of the electric field cancel out, Ey = 0.
|
|||
|
|
x -components:
|
|||
|
|
Ex = E1,x + E2,x
|
|||
|
|
Also E1,x = E2,x
|
|||
|
|
|
|||
|
|
|
|||
|
|
Electric Dipole (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
)
|
|||
|
|
n,
|
|||
|
|
q1 0
|
|||
|
|
1 y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
The y -components of the electric field cancel out, Ey = 0.
|
|||
|
|
x -components:
|
|||
|
|
Ex = E1,x + E2,x
|
|||
|
|
Also E1,x = E2,x
|
|||
|
|
Ex = 2
|
|||
|
|
( ke q
|
|||
|
|
r 2 cos θ
|
|||
|
|
)
|
|||
|
|
= 2ke q
|
|||
|
|
(a2 + y 2)
|
|||
|
|
(
|
|||
|
|
a
|
|||
|
|
√a2 + y 2
|
|||
|
|
)
|
|||
|
|
= 2ke a q
|
|||
|
|
(a2 + y 2)3/2
|
|||
|
|
|
|||
|
|
|
|||
|
|
Electric Dipole (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
0
|
|||
|
|
y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
What happens as we move infinitely far from the dipole? (y >> a)
|
|||
|
|
|
|||
|
|
|
|||
|
|
Electric Dipole (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
0
|
|||
|
|
y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
What happens as we move infinitely far from the dipole? (y >> a)
|
|||
|
|
The constant a in the denominator has less and less affect on the function. We can see that the field function approaches Efar = 2ke a q
|
|||
|
|
y3
|
|||
|
|
yli→m∞
|
|||
|
|
[E
|
|||
|
|
Efar
|
|||
|
|
]
|
|||
|
|
= yli→m∞
|
|||
|
|
|
|||
|
|
|
|||
|
|
2ke a q (a2+y 2)3/2
|
|||
|
|
2ke a q y3
|
|||
|
|
|
|||
|
|
|
|||
|
|
= yli→m∞
|
|||
|
|
|
|||
|
|
|
|||
|
|
2ke a q
|
|||
|
|
y3
|
|||
|
|
(
|
|||
|
|
(a
|
|||
|
|
y )2+1
|
|||
|
|
)3/2
|
|||
|
|
2ke a q y3
|
|||
|
|
|
|||
|
|
|
|||
|
|
=1
|
|||
|
|
|
|||
|
|
|
|||
|
|
Electric Dipole (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
0
|
|||
|
|
y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
What happens as we move infinitely far from the dipole? (y >> a)
|
|||
|
|
The constant a in the denominator has less and less affect on the function. We can see that the field function approaches Efar = 2ke a q
|
|||
|
|
y3
|
|||
|
|
yli→m∞
|
|||
|
|
[E
|
|||
|
|
Efar
|
|||
|
|
]
|
|||
|
|
= yli→m∞
|
|||
|
|
|
|||
|
|
|
|||
|
|
2ke a q (a2+y 2)3/2
|
|||
|
|
2ke a q y3
|
|||
|
|
|
|||
|
|
|
|||
|
|
= yli→m∞
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
2ke a q
|
|||
|
|
y3(
|
|||
|
|
(a
|
|||
|
|
y )2+1
|
|||
|
|
)3/2
|
|||
|
|
|
|||
|
|
|
|||
|
|
2ke a q y3
|
|||
|
|
|
|||
|
|
|
|||
|
|
=1
|
|||
|
|
|
|||
|
|
|
|||
|
|
Big-O Notation (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
0
|
|||
|
|
y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
y >> a
|
|||
|
|
Recall that f (x) = O(g (x)) if
|
|||
|
|
∣ ∣ ∣
|
|||
|
|
f (x) g (x)
|
|||
|
|
∣ ∣
|
|||
|
|
∣6C
|
|||
|
|
∀ x > k.
|
|||
|
|
|
|||
|
|
|
|||
|
|
Big-O Notation (Example 23.7)
|
|||
|
|
P
|
|||
|
|
y
|
|||
|
|
r
|
|||
|
|
a
|
|||
|
|
q a –q
|
|||
|
|
x
|
|||
|
|
u
|
|||
|
|
u
|
|||
|
|
uu
|
|||
|
|
!"
|
|||
|
|
ES
|
|||
|
|
E2
|
|||
|
|
S
|
|||
|
|
E1
|
|||
|
|
S
|
|||
|
|
0
|
|||
|
|
y 2 sin f 2 ke
|
|||
|
|
0 q2 0
|
|||
|
|
b 2 1 y 2 sin u
|
|||
|
|
y >> a
|
|||
|
|
Recall that f (x) = O(g (x)) if
|
|||
|
|
∣ ∣ ∣
|
|||
|
|
f (x) g (x)
|
|||
|
|
∣ ∣
|
|||
|
|
∣6C
|
|||
|
|
∀ x > k.
|
|||
|
|
∣ ∣ ∣ ∣
|
|||
|
|
E
|
|||
|
|
Efar
|
|||
|
|
∣ ∣ ∣ ∣
|
|||
|
|
=
|
|||
|
|
∣ ∣ ∣ ∣ ∣ ∣
|
|||
|
|
2ke a q (a2+y 2)3/2
|
|||
|
|
2ke a q y3
|
|||
|
|
∣ ∣ ∣ ∣ ∣ ∣
|
|||
|
|
=
|
|||
|
|
∣ ∣ ∣ ∣ ∣ ∣
|
|||
|
|
(
|
|||
|
|
(a
|
|||
|
|
y
|
|||
|
|
)2
|
|||
|
|
+1
|
|||
|
|
)−3/2∣
|
|||
|
|
∣ ∣ ∣ ∣ ∣
|
|||
|
|
Choosing k = a we can see:
|
|||
|
|
∣ ∣ ∣ ∣
|
|||
|
|
E
|
|||
|
|
Efar
|
|||
|
|
∣ ∣ ∣ ∣
|
|||
|
|
61
|
|||
|
|
2√2 ∀ y > a
|
|||
|
|
Therefore, E = O
|
|||
|
|
( 2ke a q y3
|
|||
|
|
)
|
|||
|
|
or simply O(y −3).
|
|||
|
|
|
|||
|
|
|
|||
|
|
Electric Dipole (Example 23.7)
|
|||
|
|
As we move away from the dipole (red line, r −3) the E-field falls off faster than it does for a point charge (blue line, r −2).
|
|||
|
|
The negative charge partially shields the effect of the positive charge and vice versa.
|
|||
|
|
|
|||
|
|
|
|||
|
|
Summary
|
|||
|
|
• electric field lines
|
|||
|
|
• the effect of fields on charges
|
|||
|
|
• the electric dipole
|
|||
|
|
Homework
|
|||
|
|
• Collected homework 1, posted online, due on Monday, Jan 22.
|
|||
|
|
Serway & Jewett:
|
|||
|
|
• NEW: Ch 23, onward from page 716. Probs: 36, 51, 61, 79
|
|||
|
|
• Understand examples 23.8 and 23.9.
|