From Infinity...to a distance R!

For those of you reading the blog, just a heads up that tomorrow IS Wednesday, our day for an AP problem quiz. Another bit of useful info...it will HEAVILY be based on kinematics. Might be a good idea to check out those review problems on kinematics handed out in class - the solutions are posted on the blog in January 2008. Use the menu on the right hand side to find those solutions....

Now for the solutions:

p. 522, 11, 12, 15, 16, 18, 20, 23

11. KE = 1/2*mv^2 with KE = 28 x 10⁶ eV * (1.6 x 10^-19 J/eV) = 4.5 x 10^-12 J. v = (2*KE/m)^(1/2) = 7.3 x 10⁷ m/s.

12. The same procedure, except we use the mass of the alpha particle. v = 1.6 x 10⁷ m/s.

15. We must first find the total potential at each location. The first location has a distance of 16 cm from both charges. Using V = kQ/R and adding the potential for each charge, the total V in the middle is 3.38 x 10⁶ V.

In the second position, the distance from one charge is 6 cm, and the other is 26 cm. We again use V = kQ/R and add the two results together to get 5.54 x 10⁶ V.

We can neglect the change of kinetic energy during this process because the charge is being 'placed' at each location.

Wncf = delta_K + delta_U = q*delta_V = 1.08 J

16.
a) The electric potential of the proton is V = ke/R = 5.8 x 10⁵ V

b) The potential energy of the system is found by multiplying the potential created by the proton (the answer from a) by the proton located at the given distance. Since V = U/q, U = q*V = 9.2 x 10^-12 J.

18. This follows the same pattern as what we did at the end of today.

The first electron takes zero work to bring together.

The second electron takes e*delta_V = -e*(k(-e)/R - 0) = ke^2/R.

The third electron must overcome the repulsive forces of BOTH electrons that are already there: e*delta_V = -e*(k(-e)/R - 0) - e*(k(-e)/R - 0)= 2ke^2/R.

The total must then be the sum of these quantities, 3*ke^2/R = 6.9 x 10^-18 J.

20. Energy is conserved, so:
initial PE + initial KE = final PE + final KE
ke*Q/R + 0 = 0 (since it is at infinity) + 1/2*mv^2
v = 2.33 x 10⁷ m/s.

23. We first need to find the total potential at each location.

At A:
V_A = kq/b + k*(-q)/(d - b) which can be simplified to kq(d - 2b)/b(d-b).

At B:
V_B = kq/(d - b) + k*(-q)/b = kq(2b - d)/b(d - b).

The difference in potential Vb - Va can be simplified to 2kq(2b - d)/b(d-b).