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: Original capacitor capacitance
\[ C_1 = \frac{A \varepsilon_0}{d} = 4\mu F \] Step 2: New capacitance with half-filled dielectric
The capacitor is treated as two capacitors in series:
\[ C_f = \frac{A\varepsilon_0}{d_1 + d_2/K} = \frac{A\varepsilon_0}{d(1 - \frac{1}{2} + \frac{1}{2K})} \] With \(K = 3\) substituted:
\[ C_f = \frac{4\mu F}{\frac{3}{2}} = 6\mu F \] The resulting capacitance is \(6\mu F\).