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Unformatted text preview: 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 Qx x 2 + a 2 ( ) 3 2 . The ring is in the xz 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 Qx a 3 x 2 a 2 + 1 3 2 1 4 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 Qx x 2 + a 2 ( ) 3 2 . The Force is of the form F = kx . Here k = Q 4 a 3 . 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 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: 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 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 E i d l a b = Q 4 1 r 2 dr a b = Q 4 1 r b 1 r a . If we use our reference point as infinity: V b V a = Q 4 1 r b 1 r a ; r a = V b = Q 4 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 j ; ! F = ! mg 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 !...
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This note was uploaded on 05/12/2010 for the course PHYSICS 1B 318007220 taught by Professor Gekelman during the Spring '10 term at UCLA.
 Spring '10
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