Question

The capacity of air filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric constant 'K' as shown in figure. The new capacity becomes $C_n$. The ratio $C_n$ to $C_0$ is ______.

• A. $\frac{K+1}{2}$
• B. $\frac{K+1}{3}$
• C. $\frac{K+1}{4}$
• D. $4(K+1)$

Hint:

Dielectric split vertically (Area divided) $\rightarrow$ Capacitors are in Parallel ($C_{eq} = C_1 + C_2$).
Dielectric split horizontally (Distance divided) $\rightarrow$ Capacitors are in Series ($1/C_{eq} = 1/C_1 + 1/C_2$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A parallel plate capacitor is half-filled with a dielectric. Based on the standard configuration implies by the options (where the dielectric fills half the area, not half the distance), we must find the new equivalent capacitance.

Step 2: Detailed Explanation:
Let the original area of the plates be $A$ and the distance between them be $d$.
The initial capacitance with air is:
$C_0 = \frac{\varepsilon_0 A}{d}$
When the dielectric fills "one-half of the space" parallel to the plates (filling half the area $A/2$ while maintaining the full distance $d$), the system effectively acts as two separate capacitors connected in parallel.
Capacitor 1 (Air half):
Area = $A/2$, Distance = $d$, Dielectric = 1
$C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{C_0}{2}$
Capacitor 2 (Dielectric half):
Area = $A/2$, Distance = $d$, Dielectric = $K$
$C_2 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{K \cdot C_0}{2}$
Since they are in parallel, the new equivalent capacitance ($C_n$) is simply their sum:
$C_n = C_1 + C_2$
$C_n = \frac{C_0}{2} + \frac{K \cdot C_0}{2}$
$C_n = C_0 \left( \frac{1 + K}{2} \right)$
We need the ratio of $C_n$ to $C_0$:
$\frac{C_n}{C_0} = \frac{K + 1}{2}$

Step 3: Final Answer:
The ratio is $\frac{K+1}{2}$, matching option (a).