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problem24_60

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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24.60: a) With the switch open: ( 29 ( 29 ( 29 F 00 . 4 1 F 6 1 F 3 1 1 F 6 1 F 3 1 eq μ C μ μ μ μ = + + + = - - C 10 8.4 V) (210 F) 00 . 4 ( 4 eq - × = = = μ V C Q total . By symmetry, each capacitor carries 4.20 C. 10 4 - × The voltages are then just calculated via V=Q/C. So: V. 70 V 70 / and V, 140 / 6 3 = - = = = = = ac ad cd ac ad V V V C Q V C Q V b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is: F. 4.5 F 6) (3 1 F 6) (3 1 1 eq μ μ μ C = + + + = - , C 10 9.5 V) (210 F) 50 . 4 ( 4 eq - × = = = μ V C Q total and each capacitor has the same potential difference of 105 V (again, by symmetry) c) The only way for the sum of the positive charge on one plate of
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