Question

Four parallel plate capacitors are connected in parallel. Each capacitor has a plate area \(\frac{A}{4}\) and the separation between the plates is \(d\), \(2d\), \(3d\) and \(4d\) respectively. The equivalent capacity of the combination is [\(\varepsilon _{0}=\) permittivity of free space]

• A. \(\frac{5A\varepsilon _{0}}{12d}\)
• B. \(\frac{7A\varepsilon _{0}}{24d}\)
• C. \(\frac{9A\varepsilon _{0}}{36d}\)
• D. \(\frac{25A\varepsilon _{0}}{48d}\)

Hint:

Capacitances add in parallel, and each capacitance is inversely proportional to its plate separation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Each capacitor has \(C=\frac{\varepsilon_{0}(A/4)}{d_{n}}\), with \(d_{n}=d,2d,3d,4d\).

Step 2: Individual capacitances
\(C_{1}=\frac{\varepsilon_{0}A}{4d}\), \(C_{2}=\frac{\varepsilon_{0}A}{8d}\), \(C_{3}=\frac{\varepsilon_{0}A}{12d}\), \(C_{4}=\frac{\varepsilon_{0}A}{16d}\).

Step 3: Parallel combination
\(C_{eq}=\frac{\varepsilon_{0}A}{4d}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\).

Step 4: Calculation
The bracket is \(\frac{25}{12}\), so \(C_{eq}=\frac{25\varepsilon_{0}A}{48d}\), option (D).

Final Answer:
The equivalent capacitance is \(\frac{25\varepsilon_{0}A}{48d}\). \[ \boxed{\text{(D) }\dfrac{25A\varepsilon_{0}}{48d}} \]