When a 6.00 V battery is connected to the plates of a capacitor, it stores a charge of 18.0 µC.?
a) What is the value of the capacitance in µF?
b) If the same capacitor is connected to a 12.00 V battery, what charge is stored?
For part (a), I got 3e-3 and web assign informed me that I was off by a multiple of ten. For part (b), I got 15.0 and web assign said that the response differs from the correct answer by more than 10%. What am I doing wrong!?
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Part a) use the equation C = Q/V
So, C = 18 x 10^-6 / 6
= 3 x 10^-6 F
b)
Using the same equation C = Q/V we can rearrange it to Q = CV
The capacitance (C) of the capacitor is 3 x 10^-6 F as we calculated in part a). V is now 12 V (it was 6 V in part a).
Q = CV
Q = 3 x 10^-6 x 12
= 36 x 10^-6 C
= 36 µC
B is 35, its an basic share 18/30 = 21/x go multiply so: 30*21/18=x (the place * ability multiply) x=35 21 is greater voltage than 18 so which you understand it might keep greater. stable good fortune.