Question 1. Answer: B Some finite time is required to change E-field Question 2. Answer: E 8x10-6C/1.6x10-19C Question 3. Answer D vy=vosin(45) and V=Ed Equate potential and KE eV=1/2 mv2 vy=sqrt(2eEd/m) solve for vo Question 4. Answer: A R=mv/qB T=2pi R/v R/v=m/qB Therefore, T=2pi m/qB The ratio of 5.00 ns to the period tells you the fraction of 2pi that is complete Question 5. Answer: C Question 6. Answer: D V=Ed PE=KE qV=1/2 mv2 qEd=1/2 mv2 E=(1/2 mv2)/qD Question 7. Answer: A Torque = u x B Question 8. Answer: B Question 9. Answer: C Find position x where E field due to both charges sums to zero. Find electric potential at position x Question 10. Answer: C UA-UB=q(VA-VB)=q(0-Ed) Question 11. Answer: B Potential energy U=kqq/r Ufinal-Uinitial

Question 12. Answer: A F=qv x B=qvBsin35 Solve for q Question 13. Answer: A Question 14. Answer: B Question 15. Answer: D Find potential due to q and Q at A and then the potential due to q and Q at B Question 16. Answer: B Question 17. Worth 10 marks A) The total energy of the alpha particle, namely, the sum of the electrical potential energy and the kinetic energy, U+K is conserved. The total energy is just the initial KE, given by K = mv2 =1.473 x 10-12 J The potential energy, U, is given by qV= (2e)V, where if we assume that the alpha particle does not penetrate the nucleus (that is, r > R), then V=kQ/r The alpha particle will slow down as it approaches the nucleus because of its increasing potential energy until all the kinetic energy is in the form of potential energy. At this point, U= initial KE U= initial KE = 1.473 x 10-12 J = (2e)(92e)k/r Solving for r we find r=2.88 x 10-14 m. Clearly the alpha particle does not reach the nuclear surface. B) For it to reach the nuclear surface, the initial KE must equal the electrical PE at the nuclear surface, UR =(2e)(92e)k/R where R= 7.4 x 10-15 m. Therefore, UR = 5.73 x 10-12 J and since UR = mv2 , then v=4.14 x 107 m/s, or more than twice the speed that it had. Question 18. Worth 20 marks (a) For a magnetic field pointing into the page, the trajectory of a positively charged particle such as a proton will be a counter-clockwise uniform circular trajectory. (b) The radius of this trajectory is R = mv/(qB) = 780 m. (c) The associated cyclotron frequency is f = 1/T = v/(2R) = 1.8 s-1. (d) The trajectory of a negatively charged particle such as an electron would be uniform and circular as well, but the particle would go clockwise (for a magnetic field pointing into the page). The m/v ratio for an electron is ~1800 times smaller than for a proton, so the radius of the trajectory would be ~1800 times smaller as well (given that the initial

velocity of the particles would be the same). The cyclotron frequency for an electron, on the other hand, would be ~1800 times higher (~3000 rotations per second).