E2HWCH23 - Chapter 23 Electrical Potential 20 The facing...

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Chapter 23 Electrical Potential 20 •• The facing surfaces of two large parallel conducting plates separated by 10.0 cm have uniform surface charge densities that are equal in magnitude but opposite in sign. The difference in potential between the plates is 500 V. ( a ) Is the positive or the negative plate at the higher potential? ( b ) What is the magnitude of the electric field between the plates? ( c ) An electron is released from rest next to the negatively charged surface. Find the work done by the electric field on the electron as the electron moves from the release point to the positive plate. Express your answer in both electron volts and joules. ( d ) What is the change in potential energy of the electron when it moves from the release point plate to the positive plate? ( e ) What is its kinetic energy when it reaches the positive plate? Picture the Problem Because the electric field is uniform, we can find its magnitude from E = V / x . We can find the work done by the electric field on the electron from the difference in potential between the plates and the charge of the electron and find the change in potential energy of the electron from the work done on it by the electric field. We can use conservation of energy to find the kinetic energy of the electron when it reaches the positive plate. ( a ) Because the electric force on a test charge is away from the positive plate and toward the negative plate, the positive plate is at the higher potential. ( b ) Express the magnitude of the electric field between the plates in terms of their separation and the potential difference between them: kV/m 5.00 m 0.100 V 500 Δ Δ = = = x V E ( c ) Relate the work done by the electric field on the electron to the difference in potential between the plates and the charge of the electron: ( 29 ( 29 J 10 01 . 8 V 0 0 5 C 10 602 . 1 Δ 17 19 - - × = × = = V q W Converting 8.01×10 - 17 J to eV yields: ( 29 eV 500 J 10 1.602 eV 1 J 10 8.01 19 17 = × × = - - W ( d ) Relate the change in potential energy of the electron to the work done on it as it moves from the negative plate to the positive plate: eV 500 - = - = W U ( e ) Apply conservation of energy eV 500 = - = U K
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to obtain: 22 •• In a potassium chloride molecule the distance between the potassium ion (K + ) and the chlorine ion (Cl ) in a potassium chloride molecule is 2.80 × 10 –10 m. ( a ) Calculate the energy (in eV) required to separate the two ions to an infinite distance apart. (Model the two ions as two point particles initially at rest.) ( b ) If twice the energy determined in Part ( a ) is actually supplied, what is the total amount of kinetic energy that the two ions have when they were an infinite distance apart? Picture the Problem
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This note was uploaded on 03/07/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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E2HWCH23 - Chapter 23 Electrical Potential 20 The facing...

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