Question

Two capacitors are connected in parallel across the terminals of a battery. One has a capacitance of 2.0 µF and the other a capacitance of 4.0 µF. These two capacitors together store 5.10 10-5 C of charge. What is the voltage of the battery?

Answer 1

When two capacitors with capacitances of 2.0 µF and 4.0 µF are connected in parallel across a battery and store a charge of 5.10 x 10^-5 C, the voltage of the battery is 8.5 volts.

Step-by-step explanation:

The question is asking to find the voltage of the battery when two capacitors with capacitances of 2.0 µF and 4.0 µF, respectively, are connected in parallel and together store a charge of 5.10 x 10-5 C. The total capacitance (Ctotal) in a parallel connection is the sum of the individual capacitances:

Ctotal = C1 + C2

Ctotal = 2.0 µF + 4.0 µF = 6.0 µF

Using the formula for charge stored in a capacitor (Q = CV), the voltage (V) across the capacitors can be found by rearranging the formula:

V = Q / Ctotal

V = 5.10 x 10-5 C / 6.0 x 10-6 F

V = 8.5 volts

Therefore, the voltage of the battery is 8.5 volts.