Question

In the figure below with V=12.0V, C1=C4=2.00, C2=4.00, and C3=1.00, what is the charge on the capacitor?

a) 6.0 C
b) 12.0 C
c) 24.0 C
d) 36.0 C

Answer 1

The charge on the capacitor is 48.0µC.

Step-by-step explanation:

In this question, we have a combination of capacitors connected in parallel. The capacitors C2 and C3 are connected in parallel and have the same potential difference of 8.0V. Therefore, the charges on these two capacitors can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.

For C2, Q2 = (2.00µF)(8.0V) = 16.0µC.

For C3, Q3 = (4.00µF)(8.0V) = 32.0µC.

Since C1 and C4 are not connected in parallel with any other capacitors, they will have the same charge as the combination of C2 and C3, which is 16.0µC + 32.0µC = 48.0µC.