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

Step 1: Understanding the Concept:
This problem requires calculating the modified capacitance of a parallel plate system after introducing a non-conducting dielectric slab that partially fills the interior space. The inserting action turns the system into a combination of two distinct capacitors linked together in series: one filled with air/vacuum and one completely filled with the dielectric material.
Step 2: Key Formula or Approach:
The general standard formula for a parallel plate capacitor containing a partial dielectric slab of thickness $t$ is: \[ C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \] Here, we substitute the specified thickness parameter value: $t = \frac{3}{4}d$.
Step 3: Detailed Explanation:
Let's place $t = \frac{3}{4}d$ into our denominator expression step by step: \[ d - t + \frac{t}{K} = d - \frac{3}{4}d + \frac{\frac{3}{4}d}{K} \] Simplify the first two terms: \[ d - \frac{3}{4}d = \frac{1}{4}d \] Now combine it with the final term over a common denominator: \[ \frac{1}{4}d + \frac{3d}{4K} = \frac{d}{4} \left( 1 + \frac{3}{K} \right) = \frac{d}{4} \left( \frac{K + 3}{K} \right) = \frac{d(K + 3)}{4K} \] Now substitute this simplified denominator back into the main capacitance formula: \[ C = \frac{\varepsilon_0 A}{\frac{d(K + 3)}{4K}} \] Invert the fraction in the denominator to find the final form: \[ C = \frac{\varepsilon_0 A}{d} \left( \frac{4K}{K + 3} \right) \] This matches option (D).
Step 4: Final Answer:
The capacitance when the slab is inserted is $\frac{\varepsilon_0 A}{d} \left( \frac{4K}{K + 3} \right)$.