Final B Solutions

# Final B Solutions - B PHYSICS 9C FINAL EXAM June 10, 2006...

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1 PHYSICS 9C FINAL EXAM June 10, 2006 LAST NAME: FIRST NAME: SECTION: Problem 1 (two points). The diagram shows two negative charges. The electric field at point P on the perpendicular bisector of the line joining them: Solution: Top charge is producing the electric field in the top left direction, while negative charge is producing the field in the bottom left direction. The resulting field is to the left. Answer D Problem 2. (two points). An electron goes from one equipotential surface to another along one of the four paths shown below. Rank the paths according to the work done by the electric field, from least to greatest. A. 4, 3, 2, 1 B. 4 and 2 tie, then 3, then 1 C. 4, 3, 1, 2 D. 1, 3, 4 and 2 tie E. 1, 2, 3, 4 Solution: The work done by electric field in cases 1,2,3,4 is W1=|e|*30 V, W2=|e|*10 V, W3=|e|*20 V, W4=|e|*10 V, where |e| is the absolute value of the charge of the electron. Therefore, W2=W4 is the least work, then W3, and W1. Answer B. B -Q -Q

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2 Problem 3. (two points) A parallel plate air capacitor has a dielectric slab partially inserted inside it. The capacitor has charge Q on its plates. Is there a force experienced by the slab? A. The capacitor is pulling the slab further inside. B. The capacitor is pushing the slab out. C. The slab is attracted to the left plate of the capacitor D. The slab is attracted to the right plate of the capacitor E. There is no force acting on the slab. Solution: The potential energy of the capacitor is a function of its capacitance, while the force acting on the slab is a derivative of the potential energy. Since the capacitor has charge Q, the potential energy of the capacitor U=Q 2 /2C, i.e. reversely proportional to its capacitance. To reach the equilibrium, the system tries to minimize its potential energy, which can be achieved by pulling the slab inside: in this case, the capacitance of the capacitor increases and leads to the decrease of the potential energy according to U=Q 2 /2C. Answer A. Problem 4. (two points) Which of the following graphs depicts the current versus voltage relationship for a device that obeys Ohm's Law? A B C D E Solution: Ohm’s law states that I(V)=V/R. Answer B. C Air ε i V i V i V i V i V +Q -Q
3 Problem 5. (two points) A proton enters a region of uniform perpendicular E and B fields. It is observed that the velocity v of the proton is unaffected. A possible explanation is: A) the given situation is impossible B) v is perpendicular to both E and B and has magnitude E/B C) v is parallel to B , so the magnetic force is zero. D) v is parallel to E and has magnitude E/B E) v is perpendicular to both E and B and has magnitude B/E Solution: Electrostatic force acting on the proton F E =q E , while magnetic force F B =q[ vB ]. To keep the proton going straight, F E = F B , or E =[ vB ]. Therefore v can be perpendicular to both

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## This note was uploaded on 11/11/2011 for the course PHY 9C taught by Professor Zieve during the Spring '08 term at UC Davis.

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Final B Solutions - B PHYSICS 9C FINAL EXAM June 10, 2006...

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