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Combination of four parallel plate air capacitors is shown in figure. Separation between the plate P1 & P2 is $\frac{d}{3}$ and separation between the plates P2 and P3 is d. 

The equivalent capacitance of the configuration is _____________$\mu$F.
(Given $\frac{\epsilon_0 A}{d} = 4\mu$F . Where A=Area of plates)

To find the equivalent capacitance of the configuration shown in the figure, follow these steps:

1. Analyze the plate arrangement:

  • The configuration is divided into two main gaps: the gap between P1 and P2 (separation $d/3$) and the gap between P2 and P3 (separation $d$).
  • Plate P1 is composed of three area segments: $A/6$, $A/3$, and $A/2$. The total area of P1 is $\frac{A}{6} + \frac{A}{3} + \frac{A}{2} = A$.
  • Middle plate P2 and bottom plate P3 also span the total area $A$.

2. Identify the series and parallel combinations:

  • The three capacitive regions between P1 and P2 are in parallel with each other. Their combined capacitance ($C_{12}$) is:
    $C_{12} = \frac{\epsilon_0 (A/6)}{d/3} + \frac{\epsilon_0 (A/3)}{d/3} + \frac{\epsilon_0 (A/2)}{d/3}$
    $C_{12} = \frac{\epsilon_0 A}{d/3} = 3 \left(\frac{\epsilon_0 A}{d}\right)$
  • The capacitive region between P2 and P3 ($C_{23}$) is:
    $C_{23} = \frac{\epsilon_0 A}{d}$
  • Since terminal 'a' is at P1 and the bottom terminal is at P3, the combined top section ($C_{12}$) and the bottom section ($C_{23}$) are in series.

3. Calculate the numerical values:

Given $\frac{\epsilon_0 A}{d} = 4\ \mu\text{F}$:

  • $C_{12} = 3 \times 4\ \mu\text{F} = 12\ \mu\text{F}$
  • $C_{23} = 4\ \mu\text{F}$

4. Calculate the equivalent capacitance ($C_{eq}$):

For two capacitors in series:
$C_{eq} = \frac{C_{12} \times C_{23}}{C_{12} + C_{23}}$
$C_{eq} = \frac{12 \times 4}{12 + 4}$
$C_{eq} = \frac{48}{16} = 3\ \mu\text{F}$

The equivalent capacitance of the configuration is 3 $\mu$F.

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Important Questions from Electricity and Magnetism

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