Question

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers \(1/3\) of the area of its plates as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is \(C_{1}\). When the capacitor is charged, the plate area covered by the dielectric gets charge \(Q_{1}\) and the rest of the area gets charge \(Q_{2}\). The electric field in the dielectric is \(E_{1}\) and that in the other portion is \(E_{2}\). Choose the correct option.

• A. \(\frac{E_{1}}{E_{2}}=4\)
• B. \(\frac{E_{1}}{E_{2}}=\frac{1}{K}\)
• C. \(\frac{C}{C_{1}}=\frac{2+K}{K}\)
• D. \(\frac{Q_{1}}{Q_{2}}=\frac{3}{K}\)

Hint:

Treat the two regions as capacitors in parallel with the same potential difference and plate separation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The slab covers one third of the plate area, so the capacitor behaves as two capacitors in parallel, with the same potential difference \(V\) and separation \(d\).

Step 2: Capacitances
\(C_1=\dfrac{K\varepsilon_0(A/3)}{d}\), and the rest \(=\dfrac{\varepsilon_0(2A/3)}{d}\), so \(C=\dfrac{\varepsilon_0A(K+2)}{3d}\).

Step 3: Ratio
\(\dfrac{C}{C_1}=\dfrac{K+2}{K}=\dfrac{2+K}{K}\), which matches option (C).

Step 4: Check the others
The field in both regions is \(V/d\), so \(\dfrac{E_1}{E_2}=1\), which rules out A and B. \(\dfrac{Q_1}{Q_2}=\dfrac{C_1}{C_2}=\dfrac{K}{2}\), not \(\dfrac{3}{K}\), so D fails.

Final Answer:
The correct relation is \(\dfrac{C}{C_1}=\dfrac{2+K}{K}\). \[ \boxed{\text{(C) }\dfrac{C}{C_1}=\dfrac{2+K}{K}} \]