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Chapter 21: Electric Fields. Honors Physics Bloom High School. 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
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Chapter 21: Electric Fields Honors Physics Bloom High School
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)
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
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 (-)?
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
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 • PhET “Charges & Fields” • PhET “Electric Field Hockey”
Section 21.1 Quiz • An electric charge produces an electric field. A test charge, q, is used to measure the strength of the field. 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?
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
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 from a neutral object making contact with a charged object
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
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
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-
Ex Prob 4How 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=Fgd/DV • 3. Solve • q=(2.4x10-14N)(0.012m)/(450V)=6.4x10-19C • q/1.60x10-19C=4e-
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
The Van de Graff Generator • Van de Graff Generator- high voltages are built up on a surface • Charges are distributed evenlyon 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
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 PhysletsI.26.1 • PhET “Capacitor Lab”
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