A 3.74 microF capacitor is connected to a 11.9 V battery. a) What is the charge

Question

A 3.74 microF capacitor is connected to a 11.9 V battery.
a) What is the charge on each plate of the capacitor? Answer in units of C. -answer: 4.4506e-5 C (inputs as correct)
b) If this same capacitor is connected to a 1.89 V battery, how much electric potential energy is stored? Answer in units of J. 002 (part 1 of 2) 10.0 points A 3.74 ?F capacitor is connected to a 11.9 V battery. What is the charge on each plate of the capacitor? Answer in units of C 003 (part 2 of 2) 10.0 points If this same capacitor is connected to a 1.89 V battery, how much electrical potential energy is stored? Answer in units of J

Explanation / Answer

the relation between charge, capacitance, and the potential difference is given by-

q=cv;

q=3.74*10^10-6 * 11.9

solving we have q=4.4*10^-5c

=4.4*e^-5c(in computer)

2) here, electric potential v=1.89 v;

now, as the same capacitor is connected the charge on the plate of the capacitor will be the same as calculated above giving c=4.4*e^-5c

now by relationship;

electrical potential energy is given by

E=v*c

=1.89*4.4*10^-5j;

=8.31*10^-5j;

= 8.31*e^-5j.

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