SS-I Physics 0175 Recitation 2 (5/24/18) Name (all group members): 1. Figure below shows a 12.0 V battery and four uncharged capacitors of capacitances C1 = 1.00 µF, C2 = 2.00 µF, C3 = 3.00 µF, and C4 = 4.00 µF. If only switch S1 is closed, what is the charge on (a) capacitor 1, (b) capacitor 2, (c) capacitor 3, and (d) capacitor 4? If both switches are closed, what is the charge on (e) capacitor 1, (f) capacitor 2, (g) capacitor 3, and (h) capacitor 4? (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: ()()()1 31 31 31.00 F 3.00 F 12.0V9.00 C.1.00 F+3.00 FC C Vq qC Cµ µµµ µ= = = =+ (b) Capacitors 2 and 4 are also in series: ()()()2 42 42 42.00 F 4.00 F 12.0V16.0 C.2.00 F 4.00 FC C Vq qC Cµ µµµ µ= = = =+ + (c) 3 19.00 C.q qµ= = (d) 4 216.0 C.q qµ= = (e) With switch 2 also closed, the potential difference V1 across C1 must equal the potential difference across C2 and is ()()3 411 2 3 43.00 F 4.00 F 12.0V8.40 V.1.00 F 2.00 F 3.00 F 4.00 FC CV VC C C Cµ µµ µ µ µ++= = =+ + + + + + Thus, q1 = C1V1 = (1.00 µF)(8.40 V) = 8.40 µC. (f) Similarly, q2 = C2V1 = (2.00 µF)(8.40 V) = 16.8 µC. (g) q3 = C3(V – V1) = (3.00 µF)(12.0 V – 8.40 V) = 10.8 µC. (h) q4 = C4(V – V1) = (4.00 µF)(12.0 V – 8.40 V) = 14.4 µC.2. Figure below shows a parallel-plate capacitor of plate area A = 10.5 cm2 and plate separation 2d = 7.12 mm. The left half of the gap is filled with material of dielectric constant κ1 = 21.0; the top of the right half is filled with material of dielectric constant κ2 = 42.0; the bottom of the right half is filled with material of dielectric constant κ3 = 58.0. What is the capacitance? Let C1 = ε0(A/2)κ1/2d = ε0Aκ1/4d, C2 = ε0(A/2)κ2/d = ε0Aκ2/2d, C3 = ε0Aκ3/2d. Note that C2 and C3 are effectively connected in series, while C1 is effectively connected in parallel with the C2-C3 combination. Thus, ()()()0 2 32 3 0 1 0 2 31 12 3 2 3 2 32 22.4 2 2 4A dC C A AC CC C d dε κ κε κ ε κ κκκ κ κ κ = + = + = + + + + With 3 21.05 10 m ,A−= × 33.56 10 m,d−= × 121.0,κ=242.0κ= and 358.0,κ= we find the capacitance to be 12 2 2 3 2113(8.85 10 C /N m )(1.05 10 m ) 2(42.0)(58.0)21.0 4.55 10 F.4(3.56 10 m) 42.0 58.0C− −−−× ⋅ × = + = × × +
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