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

Step 1: Understanding the Question:
We have a parallel plate capacitor initially with a vacuum between the plates.
A dielectric slab of thickness less than the plate separation is then inserted.
We need to find the new capacitance of this arrangement in terms of the original capacitance C and the dielectric constant K.
Step 2: Key Formula or Approach:
The initial capacitance with vacuum is \(C = \frac{\epsilon_0 A}{d}\), where A is the plate area and d is the separation.
When a dielectric slab of thickness \(t\) and dielectric constant \(K\) is inserted, the system can be viewed as two capacitors in series.
One capacitor is filled with the dielectric, with thickness \(t\), and has capacitance \(C_1 = \frac{K \epsilon_0 A}{t}\).
The other capacitor is the remaining air/vacuum gap, with thickness \((d-t)\), and has capacitance \(C_2 = \frac{\epsilon_0 A}{d-t}\).
The equivalent capacitance \(C_{eq}\) for a series combination is given by \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\).
Step 3: Detailed Explanation:
The initial capacitance is \(C = \frac{\epsilon_0 A}{d}\).
The thickness of the inserted dielectric slab is \(t = \frac{d}{3}\).
The thickness of the remaining air gap is \(d-t = d - \frac{d}{3} = \frac{2d}{3}\).
Now we find the capacitance of the two series components:
Capacitance of the dielectric part:
\[ C_1 = \frac{K \epsilon_0 A}{t} = \frac{K \epsilon_0 A}{d/3} = \frac{3K \epsilon_0 A}{d}.
\] In terms of the original capacitance C, this is \(C_1 = 3KC\).
Capacitance of the air gap part:
\[ C_2 = \frac{\epsilon_0 A}{d-t} = \frac{\epsilon_0 A}{2d/3} = \frac{3 \epsilon_0 A}{2d}.
\] In terms of the original capacitance C, this is \(C_2 = \frac{3}{2}C\).
Now, we find the equivalent capacitance for the series combination.
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3KC} + \frac{1}{(3/2)C} = \frac{1}{3KC} + \frac{2}{3C}.
\] Find a common denominator:
\[ \frac{1}{C_{eq}} = \frac{1 + 2K}{3KC}.
\] Inverting to find \(C_{eq}\):
\[ C_{eq} = \frac{3KC}{1+2K}.
\] Step 4: Final Answer:
The new capacitance of the system is \( \frac{3KC}{2K+1} \).