Question

Three parallel plate capacitors each with area \(A\) and separation \(d\) are filled with two dielectric (\(k_1\) and \(k_2\)) in the following fashion. (\(k_1>k_2\)) Which of the following is true? 

• A. \( C_B>C_C>C_A \)
• B. \( C_C>C_A>C_B \)
• C. \( C_C>C_B>C_A \)
• D. \( C_A>C_C>C_B \)

Hint:

In mixed dielectric problems, identify whether dielectrics combine in series or parallel to compare capacitances quickly.

Answer 1

The Correct Option is A

To determine which of the given sets of capacitors has the greatest capacitance, we need to analyze how the presence of dielectrics with different dielectric constants \(k_1\) and \(k_2\) affects each capacitor. Let's consider the following steps:

1. Let's define the formula for a parallel plate capacitor filled with a dielectric:

\(C = \frac{\epsilon_0 k A}{d}\), where:

• \(\epsilon_0\) is the permittivity of free space.
• \(k\) is the dielectric constant.
• \(A\) is the plate area.
• \(d\) is the separation between plates.

1. Analyze each capacitor configuration:
   • Capacitor \(C_A\): Half of the space between the plates is filled with \(k_1\) and the other half with \(k_2\). The effective capacitance for such a series combination is given by:

\(\frac{1}{C_A} = \frac{1}{C_1} + \frac{1}{C_2}\),

• where:
  • \(C_1 = \frac{\epsilon_0 k_1 A}{d/2}\)
  • \(C_2 = \frac{\epsilon_0 k_2 A}{d/2}\)
• Capacitor \(C_B\): The dielectric covers the entire plate with \(k_1\), so:

\(C_B = \frac{\epsilon_0 k_1 A}{d}\)

• Capacitor \(C_C\): Dielectric \(k_2\) is in parallel with air (or vacuum):

\(C_C = \frac{\epsilon_0 (k_2 + 1) A}{2d}\)

1. Compare the values:
   • Since \(k_1 > k_2 > 1\), the effective capacitance for \(C_A\) will be less than both \(C_B\) and \(C_C\).
   • \(C_B\) has a larger capacitance than \(C_C\), since \(C_B\) uses the greater dielectric constant \(k_1\) over the entire plate.
   • Thus, the order of capacitance is: \(C_B > C_C > C_A\).

Hence, the correct answer is \(C_B > C_C > C_A\).