Question
The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store 17 μC of charge and 5.7 x 10^-5 J of energy. When used with capacitor B, which has a capacitance of 6.9 μF, this voltage causes the capacitor to store a charge that has a magnitude of qB. Determine qB.
Answers
Damon
C = q/V
so V = q/C
so
V = qA/CA = qB/CB
and
E = (1/2) q V = (1/2) C V^2
5.7*10^-5 = (1/2)(17*10^-6)V
so
V = 6.71 volts
then
6.71 = (qB/6.9)10^6
so V = q/C
so
V = qA/CA = qB/CB
and
E = (1/2) q V = (1/2) C V^2
5.7*10^-5 = (1/2)(17*10^-6)V
so
V = 6.71 volts
then
6.71 = (qB/6.9)10^6