Phys2212_30.1+to+30.4

Phys2212_30.1+to+30.4 - Physics 2212 Electricity and...

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Physics 2212 Electricity and Magnetism Lecture 12 (Knight: 30.1 to 30.4) Calculating E from V

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10/06/09 Physics 2212 - Lecture 12 2 Chapter 29 – Summary (1)
10/06/09 Physics 2212 - Lecture 12 3 Chapter 29 – Summary (2)

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10/06/09 Physics 2212 - Lecture 12 4 The Missing Link How are E and V connected?
10/06/09 Physics 2212 - Lecture 12 5 Finding V from E q+sources U V q ( ) ; / ; / f i s s s s s U W i f F ds V U q E F q = - = - = ∆ = ( ) ( ) f i s f i s s V V S V S E ds = - = - (uniform electric field) s V E s = -

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10/06/09 Physics 2212 - Lecture 12 6 Potential from the Field of a Point Charge 2 0 ( ) ( ) 1 4 r r r V V V r q E dr dr r πε = ∞ - = - = - 2 0 0 0 1 ( ) ( ) 4 1 0 4 1 4 r r q V r V dr r q r q r = ∞ + = + - = This is the same result that we obtained in Chapter 29 from energy considerations.
10/06/09 Physics 2212 - Lecture 12 7 Example : The Potential of a Charged Disk disk 2 2 2 0 disk In Chapter 26, we found that the field on the axis of a charged disk was: 1 2 Find the corresponding potential . Q z E R z R V π ε = - + ( 29 2 2 2 0 2 2 2 2 2 2 0 0 2 2 disk 2 0 ( ) ( ) ( ) 0 1 2 2 2 This is the same expression we previously calculated. 2 z z z z z z Q z V z V E z dz dz R z R Q zdz Q dz z z R R R z R Q V z R z R = ∞ + = + - + = - = - + + = + -

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10/06/09 Physics 2212 - Lecture 12 8 Finding E from V q+sources s U W V E s q q - = = = - 0 lim s s V dV E s ds ∆ → = - = - ; ; x y z dV dV dV E E E dx dy dz = - = - = - ˆ ˆ ˆ dV dV dV E i j k dx dy dz = - + + r In other words, the E field components are determined by how much the potential V changes in the three coordinate directions. 2 0 0 1 1 For a point charge: 4 4 r dV d q q E E dr dr r r πε = = - = - =
10/06/09 Physics 2212 - Lecture 12 9 Example : The E Field of a Charged Ring ring 2 2 0 ring In Chapter 29, we found that the potential on the axis of a charged ring was: 1 4 Find the corresponding electric field . Q V z R E πε = + ( 29 3 2 2 2 0 ring 2 2 0 1 4 1 4 z dV d Q E E dz dz z R Qz E z R = = - = - + = + This is the same expression that we previously calculated.

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10/06/09 Physics 2212 - Lecture 12 10 Example : Finding E from the Slope of V An electric potential V in a particular region of space where E is parallel to the x axis is shown in the figure to the right.
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Phys2212_30.1+to+30.4 - Physics 2212 Electricity and...

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