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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) 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 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 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? 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 form 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 Milikan’s Oil Drop Experiment 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- How many electrons? • 1. Known/Unknown – Fg=2.4x10-14N, DV=450V, d=1.2cm, q=?, ne-=? • 2. Formulae – Fe=Fq – q=Fg/DV (qDV/d=Fg, solve for q) • 3. Solve – q=(2.4x10-14N)/(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 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 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 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