Question

A parallel-plate capacitor is filled with a dielectric whose dielectric constant is K, increasing its capacitance from C1 to KC1. A second capacitor with capacitance C2 is then connected in series with the first, reducing the net capacitance back to C1.

Part A
What is the capacitance C2 of the second capacitor?
Express your answer in terms of K, C1, and constants.
C2 = ?????

Answer 1

To find the capacitance C2 of the second capacitor in the series, we can use the formula C2 = C0 = C1((K1+K2)d)/(K1K2), where C1 is the initial capacitance, K1 and K2 are the dielectric constants, and d is the separation between the plates.

Step-by-step explanation:

To find the capacitance C2 of the second capacitor in the series, we first need to determine the initial capacitance C1. According to the given information, the capacitance of the parallel-plate capacitor is increased from C1 to KC1 when filled with a dielectric of dielectric constant K. Using the formula for the capacitance of a parallel plate capacitor filled with a dielectric, we can write:

C1 = (K1K2)C0/((K1+K2)d)

Where C0 represents the capacitance of the empty capacitor, K1 and K2 are the dielectric constants of the two dielectrics, and d is the separation between the plates. Rearranging the formula, we can solve for C0: C0 = C1((K1+K2)d)/(K1K2). Since the second capacitor reduces the net capacitance back to C1, we can equate C2 to C0: C2 = C0 = C1((K1+K2)d)/(K1K2).

Answer 2

The capacitance C2 of the second capacitor connected in series with the first capacitor can be calculated using the formula for capacitors in series: C2 = C1 / (K - 1)

Step-by-step explanation:

The capacitance C2 of the second capacitor connected in series with the first capacitor can be calculated using the formula for capacitors in series:

C1 = C_total = 1/C2-1/C1

Simplifying the equation:

C2 = C1 / (K - 1)