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# A capacitor of capacitance C=2.0±0.1μF is charged to a voltage V=20±0.2V. What will be the charge Q on the capacitor? Use Q=CV40±2.4×10−6 coulomb50±2.4×10−6 coulomb60±2.4×10−6 coulomb80±2.4×10−6 coulomb

A
40±2.4×106 coulomb
B
60±2.4×106 coulomb
C
50±2.4×106 coulomb
D
80±2.4×106 coulomb
Solution
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#### Q=CVC=(20±0.1)×10−6 fV=20±0.2 vQ First we find the Q of original valuesQ=2×20=40×10−6CNow Error % in C is0.12×100=5%error % in v is0.220×100=1%Total error =1+5=6%error voltage =6100×(40×10−6)=2.4×10−6∴Q=(40±2.4)×10−6C

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