Question

A slab of material of dielectric constant \(K\) has the same area \(A\) as the plates of a parallel plate capacitor, and has thickness \(\left(\frac{3}{4}d\right)\), where \(d\) is the separation of the plates. The capacitance when the slab is inserted between the plates is:

• A. \(\dfrac{\varepsilon_0A}{d}\left(\dfrac{K+3}{4K}\right)\)
• B. \(\dfrac{\varepsilon_0A}{d}\left(\dfrac{2K}{K+3}\right)\)
• C. \(\dfrac{\varepsilon_0A}{d}\left(\dfrac{K}{K+3}\right)\)
• D. \(\dfrac{\varepsilon_0A}{d}\left(\dfrac{4K}{K+3}\right)\)

Hint:

Remember: \[ d_{\text{eq}}=(d-t)+\frac{t}{K} \]

• Air gap contributes normally
• Dielectric region contributes reduced effective thickness: \[ \frac{t}{K} \]
• Smaller effective separation means larger capacitance

Answer 1

The Correct Option is D

Concept: When a dielectric slab partially fills the space between capacitor plates, the system behaves like capacitors in series. Effective separation is: \[ d_{\text{eq}}=(d-t)+\frac{t}{K} \] where: \[ t=\text{thickness of dielectric slab} \] Capacitance becomes: \[ C=\frac{\varepsilon_0 A}{(d-t)+\frac{t}{K}} \]

Step 1: Substitute slab thickness. Given: \[ t=\frac{3d}{4} \] Therefore: \[ d-t = d-\frac{3d}{4} = \frac{d}{4} \]

Step 2: Find equivalent separation. \[ d_{\text{eq}} = \frac{d}{4} + \frac{3d}{4K} \] Taking common factor: \[ d_{\text{eq}} = \frac{d}{4}\left(1+\frac{3}{K}\right) \] \[ d_{\text{eq}} = \frac{d(K+3)}{4K} \]

Step 3: Calculate capacitance. \[ C = \frac{\varepsilon_0A}{d_{\text{eq}}} \] \[ C = \frac{\varepsilon_0A}{\frac{d(K+3)}{4K}} \] \[ C = \frac{\varepsilon_0A}{d}\left(\frac{4K}{K+3}\right) \] Therefore, the correct answer is: \[ \boxed{ \frac{\varepsilon_0A}{d}\left(\frac{4K}{K+3}\right) } \]