Question

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area A/3 and separation between the plates is d, 2d and 3d respectively. The equivalent capacity of the combination is (\(ε_0\)-permittivity of free space)

• A. \(\frac{9ε_0A}{17d}\)
• B. \(\frac{11ε_0A}{17d}\)
• C. \(\frac{11ε_0A}{18d}\)
• D. \(\frac{9ε_0A}{14d}\)

Hint:

Capacitance is eps0 A / d for each; in parallel they add.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an air capacitor, \(C=\dfrac{\varepsilon_0A_{plate}}{d}\). In a parallel connection, the equivalent capacitance is the sum of the capacitances.

Step 2: Find each capacitance:
Plate area is \(\dfrac A3\). \(C_1=\dfrac{\varepsilon_0A}{3d}\), \(C_2=\dfrac{\varepsilon_0A}{6d}\), \(C_3=\dfrac{\varepsilon_0A}{9d}\).

Step 3: Add them:
\(C_{eq}=\dfrac{\varepsilon_0A}{d}\left(\dfrac13+\dfrac16+\dfrac19\right)=\dfrac{\varepsilon_0A}{d}\cdot\dfrac{6+3+2}{18}=\dfrac{11\varepsilon_0A}{18d}\). Option C.

Step 4: Why the other options are wrong.
Options A, B and D have denominators 17 or 14, which would come from adding the separations in the denominator instead of adding the capacitances.

Final Answer:
The equivalent capacitance is 11 eps0 A / (18 d). \[ \boxed{\text{(C) }\dfrac{11\varepsilon_0A}{18d}} \]