Recitation - Chapter 25

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Q 10.0x107° C P25.2 @ C= =1.00x10° F=| 1.00 uF AV 100V o A2 Q_100x10° C o C 1.00x10°F An air-filled parallel-plate capacitor has plates of area 2.30 cm? separated by 1.50 mm. (a) Find the value of its capacitance. The capacitor is connected to a 12.0-V battery. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates? e A (1.00)(8.85x10™" C*/N-m?*)(2.30x10™* m? P25.4 (a) C=——0 _ (100} LE) ) d 1.50x107° m =1.36x107"? F=|1.36 pF (b) Q=CAV =(1.36 pF)(12.0 V)=(16.3 pC AV 120V E = - = (<) 4 150x10° m 8.00x10° V/m
10. Three capacitors are connected to a battery as shown in Figure P25.10. Their capac- C EJ itances are ¢, = 3C C, = | |1 C, and C, = 5C. (a) What [ is the equivalent capacitance + of this set of capacitors? - (b) State the ranking of the capacitors according to the charge they store from largest . to smallest. (c) Rank the capac- Figure P25.10 itors according to the potential differences across them from largest to smallest. (d) What If? Assume C, is increased. Explain what happens to the charge stored by each capacitor. Cy Cs P2510 (a) Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in series with capacitor 1, so the battery +_ 3C 6C 1 1] sees capacitance | 5~ =12C|. (b) If they were initially uncharged, C, stores the same charge as C, and C, together. With greater capacitance, C, stores more charge than C,. Then | Q, >Q;>Q, |. (c) The (C,IIC;) equivalent capacitor stores the same charge as C,. Since it has greater capacitance, AV =% implies that it has smaller potential difference across it than C,. In parallel with each other, C, and C, have equal voltages: | AV, > AV, = AV, |,
(d) If C, is increased, the overall equivalent capacitance increases. More charge moves through the battery and Q increases. As AV, increases, AV, must decrease so Q, decreases. Then Q, must increase even more: | Q; and Q, increase; Q, decreases |, 26. The voltage across an air-filled parallel-plate capacitor is measured to be 85.0 V as shown in Figure P25.26a. When a dielectric is inserted and completely fills the space between the plates as in Figure P25.26b, the voltage drops to 25.0 V. (a) What is the dielectric constant of the inserted material? (b) Can you identify the dielectric? If so, what is it? (c) If the dielectric does not completely fill the space between the plates, what could you conclude about the voltage across the plates? Dielectric b Figure P25.26
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P25.26 (b) (c) (@) Note that the charge on the plates remains constant at the original value, Q,, as the dielectric is inserted. Thus, the change in the potential difference, AV =Q/C, is due to a change in capacitance alone. The ratio of the final and initial capacitances is 850V G, Q/(AV), (AV), 250V 3.40 Thus, the dielectric constant of the inserted material is Kk =3.40|, The material is probably |nylon| (see Table 25.1). The presence of a dielectric weakens the field between plates, and the weaker field, for the same charge on the plates, results in a smaller potential difference. If the dielectric only partially filled the space between the plates, the field is weakened only within the dielectric and not in the remaining air-filled space, so the potential difference would not be as small. The voltage would lie| somewhere between 25.0 V and 85.0 V.,
39. Two square plates of sides € are placed parallel to each other with separation d as suggested in Figure P25.39. You = may assume d is much less than ¢. The plates carry uni- formly distributed static charges +Q and —Q,. A block of metal has width ¢, length ¢, and thickness slightly less than d. It is inserted a distance x into the space between the plates. The charges on the plates remain uniformly dis- tributed as the block slides in. In a static situation, a metal prevents an electric field from penetrating inside it. The metal can be thought of as a perfect dielectric, with k — . (a) Calculate the stored energy in the system as a function of x. (b) Find the direction and magnitude of the force that acts on the metallic block. (c) The area of the advanc- ing front face of the block is essentially equal to €d. Consid- ering the force on the block as acting on this face, find the stress (force per area) on it. (d) Express the energy density in the electric field between the charged plates in terms of Q,, ¢, d, and €,- (¢) Explain how the answers to parts (c) and (d) compare with each other. fe——¢ [CF FEEE + 0, =X Figure P25.39
P25.39 Where the metal block and the plates overlap, the electric field between the plates is zero. The plates do not lose charge in the overlapping region, but opposite charge induced on the surfaces of the inserted portion of the block cancels the field from charge on the plates. The unfilled portion of the capacitor has capacitance &A_&!(l-x) C= d d The effective charge on this portion (the charge producing the remaining electric field between the plates) is proportional to the unblocked area: (a) The stored energy is uoQ =00 | Qhd(e-x) €T 26, 0(t-x))d | 2,F 20 2 ®) Fz_du=_i Qo ’;)d =+ Qod3 dx dx\ 2e,t 2€, 0 - 24 F= 250 e to the right | (into the capacitor: the block is pulled in) 0 _F_| & (c) Stress - 2¢, 01
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(d) The energy density is (© They are precisely the same.

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