Question

Two air-filled parallel-plate capacitors with capacitances C1 and C2 are connected in parallel to a battery that has a voltage of 34.0 V; C1 = 4.00μF and C2 = 6.00 μF.

1-What is the total positive charge stored in the two capacitors?
2-While the capacitors remain connected to the battery, a dielectric with dielectric constant 5.00 is inserted between the plates of capacitor C1, completely filling the space between them. Then what is the total positive charge stored on the two capacitors?

Answer 1

The total positive charge stored in the two capacitors before adding a dielectric is 340μC. With a dielectric inserted into C1, the charge increases to 884μC due to the increased capacitance of C1.

Step-by-step explanation:

To determine the total positive charge stored in the two capacitors connected in parallel to a 34.0 V battery, we use the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage.

• For C1 = 4.00μF, the charge Q1 = 4.00μF × 34.0 V = 136μC.
• For C2 = 6.00μF, the charge Q2 = 6.00μF × 34.0 V = 204μC.

Therefore, the total charge Qtotal before adding a dielectric is Qtotal = Q1 + Q2 = 136μC + 204μC = 340μC.

When a dielectric with an idea lectric constant of 5.00 is inserted between the plates of C1, its capacitance changes to C1' = 5 × 4.00μF = 20.00μF as C' = k × C where k is the dielectric constant. The charge on C1 with the dielectric becomes Q1' = 20.00μF × 34.0 V = 680μC, while Q2 remains 204μC.

Thus, the total charge with a dielectric is now Qtotal' = Q1' + Q2 = 680μC + 204μC = 884μC.

Answer 2

Initially, the capacitors store a total positive charge of 340.00µC. After inserting a dielectric with a constant of 5.00 into C1, the total charge stored increases to 884.00µC.

Step-by-step explanation:

When two air-filled parallel-plate capacitors with capacitances C1 and C2 are connected in parallel to a battery with a voltage of 34.0 V (where C1 = 4.00µF and C2 = 6.00µF), the total positive charge stored in the two capacitors can be calculated using the equation Q = CV. Here, C is the total capacitance in parallel which is simply the sum of the individual capacitances, and V is the voltage of the battery.

1 - The total positive charge stored on the two capacitors when connected in parallel is:

• Ctotal = C1 + C2 = 4.00µF + 6.00µF = 10.00µF
• Qtotal = Ctotal × V = 10.00µF × 34.0 V = 340.00µC

2 - After a dielectric with dielectric constant K = 5.00 is inserted into C1, the capacitance of C1 becomes C1' = K × C1. The new total charge Qtotal' considering the dielectric is:

• C1' = 5.00 × 4.00µF = 20.00µF
• Ctotal' = C1' + C2 = 20.00µF + 6.00µF = 26.00µF
• Qtotal' = Ctotal' × V = 26.00µF × 34.0 V = 884.00µC

Therefore, initially, the capacitors store 340.00µC of positive charge, and after inserting the dielectric into C1, they store 884.00µC.