Question

A parallel plate capacitor with plate area \(A\) and plate separation \(d = 2\) m has a capacitance of \(4\mu F\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K = 3\) (as shown in the figure) will be: 
 

 

• A. 2μF
• B. 32μF
• C. 6μF
• D. 8μF

Hint:

When a dielectric partially fills a capacitor, consider it as a system of capacitors in series.

Answer 1

The Correct Option is C

Step 1: {Capacitance of the original capacitor}
\[ C_1 = \frac{A \varepsilon_0}{d} = 4\mu F \] Step 2: {Finding the new capacitance when half-filled with dielectric}
The capacitor can be considered as two capacitors in series: \[ C_f = \frac{A\varepsilon_0}{d_1 + d_2/K} = \frac{A\varepsilon_0}{d(1 - \frac{1}{2} + \frac{1}{2K})} \] Substituting \(K = 3\): \[ C_f = \frac{4\mu F}{\frac{3}{2}} = 6\mu F \] Thus, the new capacitance is \(6\mu F\).

Answer 2

Step 1: Understand the original capacitor
Given:
- Plate area = \( A \)
- Plate separation = \( d = 2\,m \)
- Original capacitance \( C = 4\,\mu F \)
This is for air/vacuum between the plates (dielectric constant \( K = 1 \)).
Using the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
From this, \( \frac{\varepsilon_0 A}{2} = 4\,\mu F \)

Step 2: New setup with partial dielectric
Half the space is filled with dielectric \( K = 3 \), and the other half is air.
So the system behaves like two capacitors connected in parallel, each with plate area \( A \), but different separations:
- Capacitor 1 (with dielectric): \( d_1 = 1\,m \), \( K = 3 \)
- Capacitor 2 (without dielectric): \( d_2 = 1\,m \), \( K = 1 \)

Step 3: Calculate individual capacitances
Let’s find \( \varepsilon_0 A \) from the original setup:
\[ \frac{\varepsilon_0 A}{2} = 4\,\mu F \Rightarrow \varepsilon_0 A = 8\,\mu F \]

Now compute the two capacitances:
- With dielectric:
\[ C_1 = \frac{K \varepsilon_0 A}{d_1} = \frac{3 \times 8}{1} = 24\,\mu F \]
- Without dielectric:
\[ C_2 = \frac{\varepsilon_0 A}{d_2} = \frac{8}{1} = 8\,\mu F \]

Step 4: Combine the two capacitors
These two are in series along the vertical axis (since dielectric fills half the separation, not half the area).
So equivalent capacitance:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{24} + \frac{1}{8} = \frac{1 + 3}{24} = \frac{4}{24} \Rightarrow C_{\text{eq}} = \frac{24}{4} = 6\,\mu F \]

Final Answer: 6μF