Question

3. Show work. Two capacitors C₁ = 2.0 μF, C₂ = 4.0 μF and a 3.0 V battery are arranged as shown. When the switch S is closed the capacitors are fully charged. What is the charge on C₁? (units of μC) [ignore the red mark on the image]​

Answer 1

The charge on capacitor C₁, with a capacitance of 2.0 μF and charged by a 3.0 V battery, is calculated using the formula Q = CV, resulting in a charge of 6.0 μC.

Step-by-step explanation:

The question concerns finding the charge on capacitor C₁ when it is fully charged by a 3.0 V battery. The value of capacitor C₁ is given as 2.0 μF. To find the charge (Q) on the capacitor, we use the formula Q = CV, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitors in question are not specified to be in series or parallel, in the default scenario where each capacitor is directly connected to the battery, the voltage across C₁ is 3.0 V (the voltage of the battery). Therefore, using the formula:

Q₁ = C₁ × V = 2.0 μF × 3.0 V = 6.0 μC

This means the charge on capacitor C₁ when fully charged by a 3.0 V battery is 6.0 μC.