Princeton MCAT CONTENT REVIEW

Q1

10144 d & A - QNE= ol 79%m & Physics and M... [@ &% charge capacitor charge capacitor to volrage V to voltage V' +Q voltage = V Q then disconnect battery keep battery connected and insert dielectric and insert dielectric +Q voltage = V dielectric Vv - dielectric Q = C increases by facror of K = C increases by factor of K * Q stays the same * @ increases by factor of K V" stays the same E" Stays thc same . V decreases by a factor of K~ decreases by a factor of K . . Example 10-35: An air capacitor is charged and disconnected from the battery. The electric field between the plates is E. Now, a dielectric with dielectric constant K = 4 is inserted between the plates. What is the electric field created by the layers of induced charges on the surfaces of the dielectric? A. 3E/4 opposite the direction of E B. E/4 opposite the direction of E C. E/4 in the same direction of E D. 3E/4 in the same direction of E Solution: The question is asking for E; g,ceq- First, Ejpgueeq points in the opposite direction from E, so the answer must be either A or B. To find the magnitude of E;yy,.4, we use the fact that E - E 3. = E/K; since K = 4, we see that Eingquced = 3E/4. Thus, the answer is A. Example 10-36: A parallel-plate capacitor, with air between the plates, is charged to a voltage of V = 1000 V by a battery. The values of Q, E, and PE are also measured. The battery is then disconnected from the capacitor and a dielectric with dielectric constant K = 4 is inserted between the plates. The values of V, Q, E, and PE are measured again. Which of these values did not change? AV B. Q C E D. PE Solution: Since the battery was disconnected from the capacitor after charging, the value of Q does not change: there's nowhere for charges to go and no source of new charges. The values of V, E, and PE will all decrease by a factor of K. The answer is B. Example 10-37: A student conducts four experiments with a parallel-plate air capacitor, a battery, and a dielectric with K = 3, as described in the situations below. Which of them yielded the least amount of stored potential energy at the end of the experiment? A. Capacitor charged to voltage V by the battery, battery disconnected, dielectric inserted B. Dielectric inserted between plates of uncharged capacitor, capacitor charged to voltage V by the battery C. Capacitor charged to voltage V by the battery, battery remains connected, dielectric inserted D. Capalcitor charged to voltage V/3 by the battery, battery remains connected, dielectric inserted, capacitor charged to voltage V Solution: In choices B, C, and D, the capacitance has increased by a factor of K, and the final voltage is V. Therefore, the final potential energy is the same in all these cases; PE = lCVZ. C increases by a factor of K, and V doesn't change, so we see that PE is K times greater than it would have been if the air capacitor had been charged to voltage V. However, in choice A, we use PE = LQ2 / C and the fact that C increases by a factor of K to conclude that PE is K 2 times smaller than it would have been if the air capacitor had been charged to voltage V. Therefore, the answer i Dielectric Breakdown / Dielectrics have another purpose besides keeping the plates apart and increasing the capacitance. The il earlier for a discharging capacitor showed the plates connected by a conducting wire; the electrons on the 1. plate used this pathway to travel back to the positive plate. This type of controlled discharge is necessary if we an. tap into the motion of these electrons to do useful work (like lighting a light bulb). But why don't the extra electrons on the negative plate just jump across the gap to the positive plate, without traveling through a conducting wire |1 O