Question

Air capacitor has capacitance \(C_{1}\). The space between two plates of capacitor is filled with two dielectric as shown in figure. The new capacitance of the capacitor is \(C_{2}\). Then \(\frac{C_{1}}{C_{2}}\) is [\(d\) = distance between two plates, \(K_{1}\) and \(K_{2}\) are dielectric constants of first and second dielectrics respectively]

• A. \(\frac{11}{30}\)
• B. \(\frac{8}{30}\)
• C. \(\frac{7}{30}\)
• D. \(\frac{13}{30}\)

Hint:

The slabs sit one after the other between the plates, each of thickness d/2, so they act in series.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
From the figure the dielectrics (K1 = 5, K2 = 3) are placed side by side along the gap, each of thickness \(\dfrac{d}{2}\). They act as capacitors in series.

Step 2: Capacitances
\(C_{1}=\dfrac{\varepsilon_{0}A}{d}\). Slab capacitances: \(\dfrac{5\varepsilon_{0}A}{d/2}=\dfrac{10\varepsilon_{0}A}{d}\) and \(\dfrac{3\varepsilon_{0}A}{d/2}=\dfrac{6\varepsilon_{0}A}{d}\).

Step 3: Series
\(C_{2}=\dfrac{10\times6}{10+6}\dfrac{\varepsilon_{0}A}{d}=\dfrac{15}{4}C_{1}\), so \(\dfrac{C_{1}}{C_{2}}=\dfrac{4}{15}=\dfrac{8}{30}\).

Final Answer:
The ratio C1 to C2 is 8 over 30. \[ \boxed{\text{(B) }\dfrac{8}{30}} \]