{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

NOTES4 - Electric Potential Up to now we have developed two...

This preview shows pages 1–6. Sign up to view the full content.

2/27/03 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 generally 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 test charge q in an electric field: Defining ˆ d d = A A A , the work done by an external source to move q a distance d A in the ˆ A direction is [Joules, J] ext ext dW d q d = = − F E A A . Thus, the work required to move q from b to a is [J] a a ext ext b b W dW q d = = − E A . The force exerted by E is F E = q The external force required to hold q against E is F E ext q = − . E b a F ext F A ˆ A q

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Example : Solution: ( ) ( ) 1,1,1 0,0,1 1,1,1 0,0,1 1 1 0 0 5 ˆ ˆ ˆ ˆ ˆ ˆ 5 2 2 5 2 2 since 0 a b W q d d y x dx dy dz ydx xdy dz = − = − = − + + + + = − + = E E x y z x y z A A Along Path (1) , y x dy dx = = . Thus, choosing x to parameterize the path, 1 1 0 0 5 2 2 10[J] W xdx xdx = − + = − . Along Path (2) , y x dy xdx = = 2 2 . Thus, again choosing x to parameterize the path, Aside: For an arbitrary vector A , a b d A A depends upon the path used to get from b to a . We will show, however, that a b d E A is path independent . Given E x y z = + + 2 2 y x ± ± ± , 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) y x = 2 x z 1 1 1 y Path #2 Path #1 plane 1 z =
1 1 2 0 0 2 4 5 2 2 2 5 10[J] 3 3 W x dx x x dx = − + = − + = − . Note that in this particular problem, the work done is the same along each of the two paths.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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. Φ ( ) x is often referred to as the absolute potential or voltage . Note, however, that the line integral only defines a potential difference . Therefore, we can interpret any absolute potential at a point as the potential difference between that point and a reference point: We will define the potential difference between points a and b as the work done moving a unit charge from a to b : 0 lim [Volts, V] ( ) ( ) a ab ab q b W V d q a b = = − = Φ − Φ E A a b ab V
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 2 2 0 0 0 0 ˆ ˆ ˆ ( ) ( ) , sin 1 ˆ ˆ 4 4 1 1 1 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 19

NOTES4 - Electric Potential Up to now we have developed two...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online