Question

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness \( \frac{1}{3}d \) of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is:

• A. \( \frac{3KC}{2K+1} \)
• B. \( \frac{CK}{2+K} \)
• C. \( \frac{3CK^2}{(2K+1)^2} \)
• D. \( \frac{4KC}{3K-1} \)

Hint:

For a capacitor partially filled with a dielectric of thickness t, you can either model it as two capacitors in series or use the direct formula for effective distance: \(d_{eff} = (d - t) + \frac{t}{K}\).
The new capacitance is then \(C' = \frac{\epsilon_0 A}{d_{eff}}\).
This formula is very fast and useful to remember.

Answer 1

The Correct Option is A

The question concerns the change in capacitance of a parallel plate capacitor when a dielectric is partially inserted. Let's break this down step by step:

1. Initially, the capacitance of the capacitor with a vacuum between the plates is given by the formula: C = \frac{\varepsilon_0 A}{d} where A is the area of the plates, d is the separation between the plates, and \varepsilon_0 is the permittivity of free space.
2. When a dielectric slab of thickness \frac{d}{3} and relative permittivity K is introduced, the capacitor can be thought of as made up of two parts in series: one with the dielectric and the other with air.
3. The capacitance of the section with the dielectric slab is: C_1 = \frac{K \varepsilon_0 A}{\frac{d}{3}} = \frac{3K \varepsilon_0 A}{d}
4. The capacitance of the section without dielectric (only air) is: C_2 = \frac{\varepsilon_0 A}{\frac{2d}{3}} = \frac{3 \varepsilon_0 A}{2d}
5. Since the two sections are in series, the equivalent capacitance C'\ of the system can be found using the formula for capacitors in series: \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}
6. Substitute the values of C_1 and C_2: \frac{1}{C'} = \frac{1}{\frac{3K \varepsilon_0 A}{d}} + \frac{1}{\frac{3 \varepsilon_0 A}{2d}}
7. Simplifying gives: \frac{1}{C'} = \frac{d}{3K \varepsilon_0 A} + \frac{2d}{3 \varepsilon_0 A} \frac{1}{C'} = \frac{2d + 3Kd}{3K \varepsilon_0 A} \frac{1}{C'} = \frac{d(2K + 1)}{3K \varepsilon_0 A}
8. Rearranging gives the new capacitance C'\: C' = \frac{3K \varepsilon_0 A}{d(2K + 1)}
9. Recall the original expression for C using a vacuum: C = \frac{\varepsilon_0 A}{d}
10. Thus, substituting in terms of C we get: C' = \frac{3KC}{2K + 1}

The correct answer is therefore \frac{3KC}{2K+1}.