Unit2WEBLectureNotesNEW

Unit2WEBLectureNotesNEW - UCSD Physics 2B Summer Session I...

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UCSD Physics 2B Summer Session I Unit 2 Lecture Notes Chapter 25 Electric Potential Section 1 Energy, Work & Fields Finding the force on one charge due to one, or even several other charges was fairly straightforward. Vector sums were involved for multiple sources but, even then, the problem was at least digestible. Once we began considering continuous charge source distributions, we found it much easier to define The Electric Field q = F E and then use this to find the force on any particular charge once that field had been calculated: Force on a charge q = FE In this way we didn’t have to drag around all the source charges every time we did when using Coulomb’s force equation. Now recall the analogy made between the electric & gravitational fields. We similarly defined a gravitational field g by calculating the force on an arbitrary mass and then dividing by that very mass: m m =⇒ = F Fg g Notice that g has the units of acceleration (which happen to be 2 9.8 m/s at the surface of the Earth. In a similar way, the Electric Field E represents the force on a unit charge which, when divided by its mass, is also acceleration.
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Section 2 Electric Potential Recall that in Mechanics we defined mechanical work as force times distance, and generalized this, for non-constant force over a path, as the line integral Wd =• FL definition of work where F is the force and d L is differential length increment along the path. We used this to compute Gravitational Potential Energy of a mass using the work we do to get it there. Near the earth’s surface where the weight m g is nearly constant and d L is straight up, we got the formula P Em gh = . However : This is the negative of the work that the field itself does, since ( ) gr points down while we’re moving the mass up – we work against the field to increase PE . So let’s use the field vector g and define this potential energy as () P d =− L Note that the mass here is constant and was moved outside the integral. Since all the effort goes into computing the integral, we can define Gravitational Potential (rather than the potential energy ) by simply dividing out the mass G PE Ud m == L where ( ) is the field at the point r Notice two important things: 1. The Potential, like potential energy, is a scalar – not a vector and hence much easier to work with. 2. The Electric Potential, like its gravitational counterpart, should be path- independent since both electrostatic and gravitational fields are conservative (hint: the force equations both have the same mathematical form.) Hence we can always choose the easiest path available, say, one where E is always parallel along d L Since this looks like a vast improvement, let’s define b ab a Vd Δ= Er L Electric Potential Difference Now the work needed to move a charge from one point to the other is just the charge times the potential difference: ab ab Wq V Work to move a charge from a to b
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Units of Electric Potential Since the Electric Field E has units Force Charge Newtons Coulomb =
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Unit2WEBLectureNotesNEW - UCSD Physics 2B Summer Session I...

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