Question

A 5.00 μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. What is the charge stored in the capacitor?

Answer 1

The charge stored in a 5.00 µF capacitor charged by a 12.0 V battery is 60.00 µC, calculated using the formula Q = CV.

Step-by-step explanation:

The question involves calculating the charge stored in a parallel-plate capacitor when it is connected to a voltage source. The charge (Q) stored in a capacitor can be found using the formula Q = CV, where C is the capacitance and V is the voltage across the capacitor. In this case, the capacitor has a capacitance of 5.00 µF and is charged by a 12.0 V battery.

Using the formula, we find the charge stored in the capacitor: Q = (5.00 µF)(12.0 V) = (5.00 × 10−6 F)(12.0 V) = 60.00 µC. Therefore, the charge stored in the fully charged capacitor is 60.00 µC.