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Chapter 21: Electric Fields
Honors Physics
Bloom High School
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21.1 Creating and Measuring Electric Fields
• Electric Field- comparison to a gravitational field– Exists around any charged object (objects with mass)– Similar- acts at a distance– Dissimilar- can be positive OR negative
• Test Charge- used to determine the strength of a field and/or the direction of the field– Test charge is always positive
• Field Strength- equal to the force on the (+) test charge divided by the strength of the test charge– E=F/q (N/C)
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Relative Field Strength
Field Value (N/C)
Near a charged, hard-rubber rod 1x103
In a TV picture tube 1x105
Needed to create a spark in air 3x106
At an electron’s orbit in H 5x1011
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Practice Problem 1:Solving for Field Strength
• 1. Known/Unknown– q=5.0x10-6C, F=2.0x10-4N, E=?
• 2. Formula– E=F/q
• 3. Solve– E=(2.0x10-4N)/(5.0x10-6C)– 4.0x101N/C or 40N/C
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Practice Problem 4:Gravity vs. Electricity
• 1. Known/Unknown– Fg=2.1x10-3N, E=6.5x104N/C (down), q=?
• 2. Formulae– E=F/q (q=F/E)
• 3. Solve– q=(2.1x10-3N)/(6.5x104N/C)=3.2x10-8C– Should the charge be (+) or (-)?
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Example Problem 2
• 1. Known/Unknown– d=0.3m, q=-4.0x10-6C, E=?
• 2. Formulae– E=F/q1, F=kq1q2/d2
– Solve both for q1 and set equal to each other
– Solve new equation for E– E=kq2/d2
• 3. Solve– E=(9.0x109Nm2/C2)(-4.0x10-6C)/(0.3m)2=4.0x105N/C
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Picturing the Electric Field• Electric field is a vector
quantity– Magnitude and direction matter– Arrows extend from positive
charges and toward negative charges
– Lines closer together represent a stronger field
• Physics Physlets I.23.2, I.23.3, P.23.2
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Section 21.1 Quiz
• An electric charge, q, produces an electric field. A test charge, q, is used to measure the strength of the field generated by q. Why must q be relatively small?
• Define each variable in the formula E=F/q.• Describe how electric field lines are drawn around a
freestanding positive charge and a freestanding negative charge.
• A charge of +1.5x108C experiences a force of 0.025N to the left in an electric field. What are the magnitude and direction of the field?
• A charge of +3.4x106C is in an electric field with a strength of 5.1x105N/C. What is the force it experiences?
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21.2 Applications of Electric Fields
• Just as g=F/m describes the field strength per mass of gravity, E=F/q describes the field strength per charge– Changing the distance of either is work! (W=Fd)– Performing work on an object gives it DPE– Electric potential energy- DV=W/q (V=J/C)
• See Figure 21-5 (page 569)• Physics Physlets I.25.2
• Equipotential- when DV is zero– Moving a (+) charge around a (-) charge, keeping d
constant
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Grounding
• Charges will move until the electric PE is zero– No DV between the conductors
• Grounding- makes the electric PE between an object and the Earth 0V– Can prevent sparks resulting form a neutral object
making contact with a charged object
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DV in a Uniform Field
• By moving a charge between parallel plates, only the distance change in the field matters
• Because W=Fd and DV=Fd/q– DV=Ed– The potential difference is
equal to the field strength multiplied by the distance the charge is moved
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Practice Problem 16
• 1. Known/Unknown– E=6000N/C, d=0.05m, DV=?
• 2. Formula– DV=Ed
• 3. Solve– DV=(6000N/C)(0.05m)=300V
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Milikan’s Oil Drop Experiment
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Oil Drop Rationale
• If a known electric field is applied to the plates (F=E/q) and the mass is found of each droplet (F=mg), the charge can be found for a single droplet!– mg=Eq q=mg/E
• The charges were found to always be multiples of 1.60x10-19C, which we now know is the charge of a e-
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How many electrons?
• 1. Known/Unknown– Fg=2.4x10-14N, DV=450V, d=1.2cm, q=?, ne-=?
• 2. Formulae– Fe=Fq (qDV/d=Fg, solve for q)
– q=Fg/DV
• 3. Solve– q=(2.4x10-14N)/(450V)=6.4x10-19C– q/1.60x10-19C=4e-
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Sharing Charges
• All systems desire equilibrium– Charges we distribute
themselves evenly across any available surface
– When 2 spheres touch, they act as a single object
– Charge to area ratio is what counts
– Charge density the greatest near points
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The Van de Graff Generator
• Van de Graff Generator- high voltages are built up on a surface– Charges are distributed evenly on surface– Very large charge is possible (MV range!)– Ours builds to 750,000V
• MythBusters: Van de Graff Generator– http://www.youtube.com/watch?v=7qgM1A3pgkQ
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Storing Charges:The Capacitor
• Capacitor- stores electrical charge– Used in all circuitry– Storage based on voltage, size
of plates and gap between plates
• Capacitance (C)- ratio of charge stored to electric potential difference– C=q/DV– Measured in Farads (F=C/V)– Typically 10-12 to 10-6 F
• Physics Physlets I.26.1
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Practice Problem 31
• 1. Known/Unknown– C1=3.3mF, C2=6.8mF, DV=24V, q1=?, q2=?
• 2. Formula– C=q/DV q=DVC
• 3. Solve– q1=(24V)(3.3x10-6F)=7.92x10-5C
– q2=(24V)(6.8x10-6F)=1.63x10-4C