Question

Two parallel plate air capacitors are connected in parallel. Each capacitor has plate area \(A/2\) and separation between the plates is \(d\) and \(2d\) respectively. The equivalent capacity of the combination is

( \(ε_0\) = absolute permittivity of free space)

• A. \(\frac{Aε_0}{d}\)
• B. \(\frac{3Aε_0}{4d}\)
• C. \(\frac{2Aε_0}{3d}\)
• D. \(\frac{Aε_0}{4d}\)

Hint:

Add the two capacitances, each equal to epsilon0 times area over separation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A parallel plate air capacitor has \(C = \dfrac{\varepsilon_0 A_{plate}}{d}\). Capacitors connected in parallel add up.

Step 2: Compute
\[ C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0A}{2d},\qquad C_2 = \frac{\varepsilon_0 (A/2)}{2d} = \frac{\varepsilon_0A}{4d} \]
\[ C = C_1 + C_2 = \frac{\varepsilon_0A}{2d} + \frac{\varepsilon_0A}{4d} = \frac{3\varepsilon_0A}{4d} \]
Option (D) is only \(C_2\), and (A) would need both capacitors to be as large as \(C_1\) twice.

Final Answer:
The equivalent capacity is \(\dfrac{3A\varepsilon_0}{4d}\), option (B). \[ \boxed{\frac{3A\varepsilon_0}{4d}} \]