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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