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10/10/01 Electric Potential Up to now we have developed two methods for determining the E field from static charge distributions: Poisson's integral for E in a homogenous medium, which was developed by generalizing Coulomb's law. This integral can be very difficult to evaluate analytically. Gauss' law, which is relatively simple to apply but it requires charge distributions that vary only along a single coordinate. In this next section we will develop another technique that involves introducing an auxiliary “ scalar potential ” function. We will find that The scalar potential is generaly easier to compute from a given charge distribution than the E field. The E field can be computed from the scalar potential. To begin, assume that we place a text charge q in an electric field The external force required to hold q against E is FE ext q = − . Defining dd AA A = ± , the work done by an external source to move q a distance d A in the ± A direction is dW d q d ext ext =⋅ = [Joules, J] . Thus, the work required to move q from b to a is [J] aa ext ext bb Wd Wq d == ∫∫ E A . The force exerted by E is = q Aside: For an arbitrary vector A , A z d b a A depends upon the path used to get from b to a . We will show, however, that E z d b a A is path independent . This set of notes is related to the material in Sections 3.5, 2.5, 2.8, and 2.9 in the text by D.K. Cheng

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2 Example : Solution: () ( ) 1,1,1 [C] 0,0,1 0,0,1 11 00 5 ˆˆ ˆ ˆˆˆ 52 2 2 s i n c e 0 a b Wqd d y xd x d y d z ydx xdy dz =− + + + + ⎡⎤ + = ⎢⎥ ⎣⎦ ∫∫ EE x y zx y z AA Along Path (1) , y x dy dx =⇒ = . Thus, choosing x to parameterize the path, Wx d x x d x + L N M O Q P z z 2 1 0 0 1 0 1 [J]. Along Path (2) , yx d y x d x =⇒= 2 2 . Thus, again choosing x to parameterize the path, W x dx x xdx + L N M O Q P + L N M O Q P z z 2 2 5 2 3 4 3 10 2 0 1 0 1 [[ C] C] Note that in this particular problem, the work done is the same along each of the two paths. Given Exy z =++ 22 ±± ± , find the work required to move a 5 [C] charge from (0,0,1) to (1,1,1) along paths (1) and (2). (1) y x = (2) = 2
3 Definition: Potential Difference In this expression Φ () x is the potential at x relative to a reference point where the potential is defined to be zero ( ground ). Φ x is often referred to as the absolute potential or voltage . Note, however, that we can only define potential difference. Therefore, any absolute potential at a point is always the potential difference between that point and the reference point. We will define the potential difference between points a and b as the work done per unit charge, V W q d ab ab q ab b a == =− z lim 0 E A [Volts, V] ΦΦ

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4 Example: Point charge at the origin The potential difference between the spherical surface with a radius a and a spherical surface with a radius b is ΦΦ () ±± .
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## notes4 - This set of notes is related to the material in...

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