1Bweek6. - Physics 1B Sixth Week Class Notes Spring 2010...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Physics 1B Class Notes © Walter Gekelman Sixth Week Spring 2010 Electrical Potential Finally let us do back to a ring of charge. We derived an equation for the electric field along the x axis of the ring E x = 1 4 πε 0 Qx x 2 + a 2 ( ) 3 2 . The ring is in the x-z plane. Let us consider the motion of a negative particle released at a small value of x (very close to the center of the ring). In this case x <<a : E x = 1 4 0 Qx a 3 x 2 a 2 + 1 3 2 1 4 0 Qx a 3 1 3 2 x 2 a 2 . Since x << a we can neglect the second term in the series and explicitly putting Q < 0 E x = 1 4 0 Qx x 2 + a 2 ( ) 3 2 . The Force is of the form F = kx . Here k = Q 4 π a 3 ε 0 . This will behave just like a spring does and the charge will oscillate back and forth about the center of the ring. We can associate a force with a potential : F = dU dx or U = 1 2 Q 4 a 3 0 x 2 . This is a parabolic potential energy diagram or The same concept we used for the spring applies to the electron! The definition of the electrical potential is:
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 (1) U ba = U b U a = E i d l a b . Notice this is the potential difference between two points a and b. There is a dot product in the integral which means that at each position of the path between a and b only the component of E along the path matters. This makes the potential difference independent of the path chosen to get from a to b. There is an analogy between the electric potential and the work that has to be done raising a mass in a gravitational field. This is shown in the diagram below. The only difference is that the electric potential difference is the work moving a unit or test charge and q does not enter the expression as m does in the work against gravity. a start b end
Background image of page 2
3 What is the potential of a single charge Q. It is discussed in the text but to make it clearer consider moving a test charge from point b near charge Q out to infinity (point a) The potential difference is V ba = V b V a = Q 4 πε 0 E i d l a b = Q 4 0 1 r 2 dr a b = Q 4 0 1 r b 1 r a . If we use our reference point as infinity: V b V a = Q 4 0 1 r b 1 r a ; r a = V b = Q 4 0 1 r b Gravitational Field Electric Field Work against gravity = mgh Work moving test charge W = ! ! F i d ! l " in raising the mass d ! l = dy ˆ j ; ! F = ! mg ˆ j W = mgdy a b " = mg ( y b ! y a ) U b ! U a " W q = ! ! E i d ! l a b # to raise the charge from - to + d ! l=dy ˆ j ; ! E = ! F q = ! E ˆ j = $ % 0 ! ˆ j ( ) U = dy 0 = 0 a b # y b ! y a ( ) In this case we are moving outwards b a dl Q dl !
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/12/2010 for the course PHYSICS 1B 318007241 taught by Professor Corbin during the Spring '10 term at UCLA.

Page1 / 16

1Bweek6. - Physics 1B Sixth Week Class Notes Spring 2010...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online