Rankers Physics
Topic: Capacitors
Subtopic: Parallel Plate Capacitor

Plates of area A are arranged as shown. The distance between each plate is d, the net capacitance is : Image related to
\[\frac{\varepsilon_{0}A}{d}\]
\[\frac{7\varepsilon_{0}A}{d}\]
\[\frac{6\varepsilon_{0}A}{d}\]
\[\frac{5\varepsilon_{0}A}{d}\]

Solution:

The arrangement appears to be a system of parallel plates connected alternately to terminals aa and bb. Let’s determine the net capacitance.

Key Observations:

  1. The plates form a series-parallel combination.
  2. The area of each plate is AA, and the separation between adjacent plates is dd.
  3. The effective configuration can be reduced to find the equivalent capacitance.

Equivalent Capacitance Derivation:

  1. Pairing of Plates:
    • Adjacent plates (connected alternately) act as capacitors.
    • Each capacitor has a capacitance C=ε0AdC = \frac{\varepsilon_0 A}{d}.
  2. Parallel and Series Combination:
    • There are three capacitors in the arrangement, effectively forming a single network.
    • The middle plate shares equal charge with both sides, simplifying to an equivalent capacitance of ε0Ad\frac{\varepsilon_0 A}{d}.

Thus, the net capacitance is:

Cnet=ε0Ad.C_{\text{net}} = \frac{\varepsilon_0 A}{d}.

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