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# 1Bweek6 - 1 Physics 1B Class Notes © Walter Gekelman Sixth...

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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 πε 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 πε Qx a 3 x 2 a 2 + 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 3 2 ≅ 1 4 πε 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 πε − Qx x 2 + a 2 ( ) 3 2 . The Force is of the form F = − kx . Here k = − Q 4 π a 3 ε . 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 πε 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 πε E i d l a b ∫ = − Q 4 πε 1 r 2 dr a b ∫ = Q 4 πε 1 r b − 1 r a ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . If we use our reference point as infinity: V b − V a = Q 4 πε 1 r b − 1 r a ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ; r a = ∞ V b = Q 4 πε 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 !...
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1Bweek6 - 1 Physics 1B Class Notes © Walter Gekelman Sixth...

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