P77 A capacitor consists of two rectangular metal plates 3 m by 4 m, placed a di
ID: 1580517 • Letter: P
Question
P77 A capacitor consists of two rectangular metal plates 3 m by 4 m, placed a distance 2.5 mm apart in air (Figure 19.85). The capacitor is connected to a 9 V power supply long enough to charge the capacitor fully, and then the battery is removed. 2.5 mm 4 m not to scale): 1 Voltmeter 1 1 mm Figure 19.85 (a) Show that there will not be a spark in the air between the plates. (b) How much charge is on the positive plate of the capacitor? With the battery still disconnected, you insert a slab of plastic 3 m by 4 m by 1 mm between the plates, next to the positive plate, as shown in Figure 19.85. This plastic has a dielectric constant of 5. (c) After inserting the plastic, you connect a voltmeter to the capacitor. What is the initial reading of the voltmeter? (d) The voltmeter has a resistance of 100 MS2 (1 × 10° S). What does the voltmeter read 3 s after being connected?Explanation / Answer
(A) field in between plates, E = V/d
E = 9 V / (2.5 x 10^-3) = 3600 V/m
that is vey much less that field strength of air,
hence there will be no spark.
(B) Q = CV and C = e0 A / d
C = (8.854 x 10^-12)(3 x 4) / (2.5 x10^-3)
C = 42.5 x 10^-9 F
Q = 382.5 x 10^-9 C ....Ans
(C) now its like two capacitors are connected in series,
C1 = k e0 A /d1 = (8.854 x 10^-12 x 5 x 3 x 4)/(1 x 10^-3)
C1 =531.2 x 10^-9 F
C2 = e0 A /d2 = (8.854 x 10^-12 x 3 x 4)/(1.5 x 10^-3)
C2 = 70.8 x 10^-9 F
C = C1 C2 / (C1 + C2) = 62.5 nF
V = Q / C = 382.5 x 10^-9 / 62.5 x 10^-9 = 6.12 Volt
(D) T = R C = (62.5 x 10^-9 x 1 x 10^8) = 6.25 sec
Vc = 6.25[ e^(-t/T)] = 6.25 [ e^(-3/6.25)]
Vc = 3.87 Volt
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