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Quiz 3 |
| * DISCUSSIONS PROVIDED |
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| READ ALL EXAMPLES ! SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. |
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Quiz 3 |
| * DISCUSSIONS PROVIDED either in class or on the web. |
| DISCUSSIONS SOON |
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4. See example 1. We know the potential V at a distance r from a
single point charge is k*q/r,. When another point charge Q
placed at that location, the potential energy created is U = Q*V =
KQ*q/r. Note: Change in U = -W, where W is the work by the
field required to bring Q from infinity to a distance r from charge
q. We want to compute the work W(a to b) done by the field when the distance between Q and q is changed from ra to rb. W(a to b) = -change in U. Now, change in U = k*Q*q/rb - k*Q*q/ra. Thus: W(a to b) = -change in U = k*Q*q/ra - k*Q*q/rb . Since the second radius ( at b) is smaller, the work by field W(a to b) < 0. Note: In moving the charge from a to b, the work Wyou you do is -W(a to b). Thus Wyou > 0. Note work you do Wyou = change in U. (b) Use conservation of energy (Ch. 7); KE means kinetic energy, U means potential energy, point 1 is when the two charges are close together, and point 2 is when they have moved back to the original distance : KE1 + U1 = KE2 + U2. Assume KE1 = 0 since they start from rest. Thus KE2 = U1 - U2 = where U1 is the potential energy when the protons are 3.00 x10-15 m apart and U2 is when they are 2.00x10-10 m apart. Thus KE2 > 0. |
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5. (a) KEi + Ui = KEf + Uf. See section 23.2: U =
charge*V. We consider the potential to be due to source
charge q1, the stationary (at rest) charge. KEi = initial kinetic energy of charge 2, the only charge that's moving in the problem. Ui = = q2*Vi = k*q2*q1/r i . Note ri = 0.800 m. KEf = final KE of charge 2. Uf =q2*Vf = k*q2*q1/rf . Note: Note rf = 0.400 m. Find KEf and the final speed of charge 2. (b) Use the initial kinetic and potential energy given in the problem at separation distance ri = 0.800 m. KEi + Ui = KEf + Uf. In this case KEf = 0 for charge 2. Find Uf and rf where charge 2 comes to rest. |
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12. Assume the particles (protons) are separated by an infinite
distance when they start out. They are initially headed toward
each other at speed v = 1000 km/s = 1000000 m/s. (Lower case v
denotes speed.) KEi + Ui = KEf + Uf. Now, U = k*q1*q2/r , from page 759 dealing with the electric potential energy of several point charges. In this case there are two charges. KEi = 2*(1/2)m*v2, where initial speed v = 1000 km/s (convert to m/s!) is given for protons. Ui = 0 since ri is essentially infinite. KEf = 0 when they come to rest momentarily at minimum separation and turn around. Uf = k*e2/rf . Find the final separation. |
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14. Use work by field = -change in U . Now, change in U = q*change in V Change in V = -E*(6.00 m) . Thus, work by field = -change in U = -change in V = + E*(6.00 m). The work by field is positive as you would expect for motion of a positive charge in the direction of the electric field. (a) KEi + Ui = KEf + Uf. Thus change in U = KEi - KEf and thus work by field = -change in U = KEf - KEi, where the initial kinetic energy is zero and the final KE is given. (b) change in V = change in U/q = (KEi - KEf)/q (c) E*(6.00 m) = -change in U. |
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16. Here is a schematic of the arrangement of the two stationary,
positive point charges: (Q1)----------------(Q2) The two charges above are separated by distance d. The electron
starts at the center and is more attracted to the charge on the right
end than the charge on the left end since we assume the right charge Q2
= +3.00 nC (nano -coulomb); the electron moves rightward. |
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