Chap_20_soln[1]

Chap_20_soln[1] - 20-16: Picture the Problem: The kinetic...

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20-16: Picture the Problem : The kinetic energy of an electron changes as it moves to locations of different electric potential. Strategy: As the electron moves from place to place, its kinetic energy increases as it moves toward higher potential (remember, it is negatively charged, so it is accelerated opposite the field and toward positive charges) and decreases as it moves toward lower potential. Use the conservation of energy to determine the potential at point C and the kinetic energy at point A. Solution: 1. (a) Set ii ff K UKU  for the first case: AA 0 UK U C  2. Now set K   for the second case: BA 02 U C 3. Solve the first equation for AAC K UU  U and substitute into the second, and solve for C :   C CA B 2 2 U U U C 4. Divide both sides by to find e C : V      B B 2 2 2 332 V 149 V 515 V eV eV eV VV V  5. (b) Substitute the result from part (a) into the expression from step 1 to find A : K   A C 332 515 V 183 eV KUU e V V e  Insight: In the first case the electron moves from 332 V (point A) to 515 V (point C), gaining 183 eV of kinetic energy, but in the second case the electron moves from 149 V (point B) to 515 V (point C), gaining 366 eV of kinetic energy. 20-20: Picture the Problem : A proton accelerates from rest through a large potential difference. Strategy: The change in potential energy of the proton is the charge times the potential difference (equation 20-2). The change in potential energy equals the gain in kinetic energy, which can then be used to find the potential difference. Solution: 1. Set the kinetic energy equal to the change in potential energy and solve for v : 2 2 1 2 2 mv mv U e V V e       2. Find for : V 0.1 vc   2 2 2 27 8 2 19 1.67 10 kg 0.1 3 10 m/s 0.1 5 MV 22 21 .6 10 C mc mv V ee   Insight: It takes a much higher potential difference to accelerate a proton to this speed than it does to accelerate an electron, which would reach 10% of the speed of light with a potential difference of only 2.5 kV. 20-28: Picture the Problem : The nucleus at the origin creates an electric potential in the surrounding region.
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This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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Chap_20_soln[1] - 20-16: Picture the Problem: The kinetic...

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