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Understanding Electric Fields: Creation, Measurement, and Applications

This chapter explores the fundamentals of electric fields in physics, comparing them to gravitational fields and discussing their properties. It details the concept of a test charge used to measure electric field strength, provides equations related to electric fields, and demonstrates their applications in real-world scenarios like Milikan's oil drop experiment. Additionally, it covers how electric potential energy relates to electric fields, grounding mechanisms, and capacitors. Through practice problems, it illustrates the calculations involved in determining field strength and charge relationships.

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Understanding Electric Fields: Creation, Measurement, and Applications

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  1. Chapter 21: Electric Fields Honors Physics Bloom High School

  2. 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)

  3. Relative Field Strength

  4. 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

  5. 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 (-)?

  6. 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

  7. 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

  8. 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?

  9. 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

  10. 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

  11. 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

  12. 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

  13. Milikan’s Oil Drop Experiment

  14. 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-

  15. 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-

  16. 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

  17. 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

  18. 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

  19. 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

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