Question

A parallel plate capacitor has capacitance $C$. If a dielectric slab of dielectric constant $K$ and thickness equal to half the plate separation is introduced parallel to the plates, the new capacitance is:

• A. $\dfrac{2K}{K+1}\,C$
• B. $\dfrac{K+1}{2K}\,C$
• C. $\dfrac{K}{K+1}\,C$
• D. $KC$

Hint:

Think of the system as two capacitors in series: $C_{air} = \dfrac{\varepsilon_0 A}{d/2}$ and $C_{dielectric} = \dfrac{K\varepsilon_0 A}{d/2}$. Their series combination gives $\dfrac{2K}{K+1}C$.

Answer 1

The Correct Option is A

Step 1: Concept
A partial dielectric divides the capacitor into two capacitors in series: one air gap and one dielectric-filled gap.

Step 2: Meaning
With plate separation $d$ and slab thickness $t = d/2$, the capacitance formula is $C' = \dfrac{\varepsilon_0 A}{d - t + t/K}$.

Step 3: Analysis
Substituting $t = d/2$: \[C' = \frac{\varepsilon_0 A}{d - \tfrac{d}{2} + \tfrac{d}{2K}} = \frac{\varepsilon_0 A}{\tfrac{d}{2}+\tfrac{d}{2K}} = \frac{\varepsilon_0 A}{\tfrac{d(K+1)}{2K}} = \frac{\varepsilon_0 A}{d}\cdot\frac{2K}{K+1} = C\cdot\frac{2K}{K+1}.\]

Step 4: Conclusion
The new capacitance is $C' = \dfrac{2K}{K+1}\,C$. Final Answer: (A)