Question

The Figure below shows a system of four capacitors, where the potential difference across ab is 50.0 V .

A. Find the equivalent capacitance of this system between a and b.
B. How much charge is stored by this combination of capacitors?
C. How much charge is stored in the 10.0-μF capacitor?
D. How much charge is stored in the 9.0-μF capacitor?

Answer 1

The equivalent capacitance of the system between points a and b is 50.0 μF. The combination of capacitors stores a charge of 2500.0 μC. The 10.0 μF capacitor stores a charge of 500.0 μC, and the 9.0 μF capacitor stores a charge of 450.0 μC.

Step-by-step explanation:

To find the equivalent capacitance of the system between points a and b, we first need to determine the total capacitance of the capacitors connected in parallel. The formula for the total capacitance of capacitors in parallel is:

C_total = C1 + C2 + C3 + C4

Plugging in the values given in the diagram, we have:

C_total = 20.0 μF + 15.0 μF + 10.0 μF + 5.0 μF = 50.0 μF

The charge stored by this combination of capacitors can be determined using the formula:

Q = C_total * V

Plugging in the values from the diagram, we have:

Q = 50.0 μF * 50.0 V = 2500.0 μC

The charge stored in the 10.0 μF capacitor can be determined using the formula:

Q = C * V

Plugging in the values from the diagram, we have:

Q = 10.0 μF * 50.0 V = 500.0 μC

The charge stored in the 9.0 μF capacitor can be determined using the same formula:

Q = 9.0 μF * 50.0 V = 450.0 μC