Download PHY-102: Electrostatics – Lecture 4 Electric Potential and more Slides Thermodynamics in PDF only on Docsity! PHY-102: ELECTROSTATICS– LECTURE 4 ELECTRIC POTENTIAL LECTURE OUTLINE • Work Done in Electric Field • Electric Potential Energy • Electric Potential • Equipotential Surfaces • Finding V from E • Potential due to Electric Dipole • Calculating E from V • Potential of a charged conductor POTENTIAL ENERGY AND WORK DONE ➢ Electric field is a conservative field. ➢ Due to work done there is a change in potential energy of the charge field system given by U = 𝑈𝑓 −𝑈𝑖 =−WAB U = −q0Es cos B A s E C
POTENTIAL ENERGY AND WORK DONE :
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POTENTIAL ENERGY AND WORK DONE • Cases • ∆𝑈 > 0 • ∆𝑈 = 0 • ∆𝑈 < WORK: POSITIVE OR NEGATIVE? 2. In the right figure, we move the proton from point i to point f in a uniform electric field directed as shown. Which statement of the following is true? A. Electric field does positive work on the proton; And Electric potential energy of the proton increases. B. Electric field does negative work on the proton; And Electric potential energy of the proton decreases. C. Our force does positive work on the proton; And Electric potential energy of the proton increases. D. Electric field does negative work on the proton; And Electric potential energy of the proton decreases. Eif ELECTRIC POTENTIAL FROM ELECTRIC FIELD −= f i sdE q U V 0 October 3, 2007 ❑ The electric potential energy ◼ Start ◼ Then ◼ So ❑ The electric potential q U V = q U q U q U VVV if if =−=−= sdFdW = sdEqdW = 0 sdEqW f i = 0 −=−=−= f i if sdEqWUUU 0 ❑ Potential difference depends only on the source charge distribution (Consider points i and f without the presence of the test charge; ❑ The difference in potential energy exists only if a test charge is moved between the points. POTENTIAL DUE TO A POINT CHARGE • Start with (set Vf=0 at and Vi=V at R) • We have • Then • So • A positively charged particle produces a positive electric potential. • A negatively charged particle produces a negative electric potential 2 04 1 r q E = 2 04 1 r q E = −=−=−=−= f i R f i if EdrdsEsdEVVV )0cos( r q rV 04 1 )( = R q r q dr r q V R R 00 2 0 4 11 4 1 4 0 −= =−=− END PROBLEMS -6.00 0
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28. Each side of the square has length a.
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er the charges clockwise
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trom the upper left hand corner. Then fe __
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The amount of work required el
