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
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.
1 answer
1 answer