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### Lect4-ch17__2

Course: PHY 2054, Fall 2009
School: Fayetteville State...
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2054C L4Ch17 PHY College Physics B Summer 2006 Electricity, Magnetism, Light, Optics and Modern Physics Dr. David M. Lind Todays Lecture: purpose &amp; goals To understand: Electric Potential Energy Electric Potential 3) Equipotential Lines 4) Capacitance &amp; dielectrics Why do we care? We will find that it is by the control of the motion of electrical charges by manipulation of electrical...

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2054C L4Ch17 PHY College Physics B Summer 2006 Electricity, Magnetism, Light, Optics and Modern Physics Dr. David M. Lind Todays Lecture: purpose & goals To understand: Electric Potential Energy Electric Potential 3) Equipotential Lines 4) Capacitance & dielectrics Why do we care? We will find that it is by the control of the motion of electrical charges by manipulation of electrical potentials that make possible many of the luxuries we use each day CDs, DVDs, plasma TVs, radios, cell-phones, any kind of electronics. Equipotential Lines and Field Lines Equipotential Lines = lines of constant potential Constants may be given in V Field lines equipotential lines Capacitance We have seen that charges possess PE in an electric field. A device that stores energy by locating charges in a field is called a Capacitor The battery provides work by bringing to and + to + The more charge it stores at a given voltage, the more energy is stored Capacitance C U The total electrical energy stored in a capacitor is Q V Unit 1 F Farad= 1 Coulomb 1 Volt 1 QV 2 1 CV 2 2 1 Q2 2C Parallel Plate Capacitor For parallel plate capacitors; ++A +C0 ++d +2 +1 12 C + 8.85 10 +2 0 4k Nm +++Why is that so ? Two factors determine charge on plates: Repulsion between like charges on either plate larger Area increases capacitance Attraction between unlike charges smaller distance increases capacitance Dielectric Materials A B C Dielectric materials contain electric dipoles which align to the field. => keeping charge constant, but reducing field E in the material (acts like large deduction in distance) => V = E d, reduction of V => increase in C Original: C C 0 New (with dielectric): A d 0 k A d Question A parallel-plate capacitor is attached to a battery that maintains a constant potential difference V between the plates. The battery then is disconnected and a glass slab is inserted so as to just fill the space between the plates. When the glass is inserted, The 1. voltage between the capacitor plates is increased 2. The voltage stays the same 3. The voltage decreases Problem solving Coulombs Law Q1 Q2 Fk2 r k 8.988 10 9 2 Nm C2 EXAMPLE 16-9 to determine the electrical force one charge from multiple other charges: i. Calculate the magnitude of force from each of other charges using Coulombs law ii. Determine direction of force using the same/different rule. iii. Draw a careful vector diagram showing what is happening to the object. iv. Examine the x and y-contributions separately. v. Resolve each vector into x- and ycomponents vi. Add all the x-components total (resultant) x-component vii. Add all the y-components total (resultant) y-component viii. Determine the magnitude and direction of the final resultant vector (convert back from component to vector form). Some comments about tests. . . . . . . . and problem solving the Mini-exams similar to CAPA and recommended problems in the book. Recommend that you look at all of them. Hint! Hint! Hint! Test Etiquette: -- Bring a pencil, your student ID, and a calculator; Leave your cell phone at home. -- You will not need any other materials than those listed above. (equations and nu...

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