Question

Four metal plates, each with surfa area $A$ on one side, are plad with separation $d$ as shown in the figure. The capacitan between a and b is ($\varepsilon_0$ -- permittivity of free spa)

• A. $\dfrac{3\varepsilon_0 A}{d}$
• B. $\dfrac{2\varepsilon_0 A}{d}$
• C. $\dfrac{2\varepsilon_0 A}{3d}$
• D. $\dfrac{3\varepsilon_0 A}{2d}$

Hint:

Capacitance with Multiple Plates:

• Capacitance of a pair: $C = \dfrac\varepsilon_0 Ad$
• Series: $\dfrac1C_eq = \sum \dfrac1C_i$
• Parallel: $C_eq = \sum C_i$
• Analyze electrical connectivity carefully to assign series/parallel.

Answer 1

The Correct Option is C

The capacitance between parallel plates is given by $C_0 = \dfrac{\varepsilon_0 A}{d}$.
Let the configuration be such that Plate 1 and 3 are connected to terminal a, and Plates 2 and 4 to terminal b.
This forms three capacitors: $C_{12}$, $C_{23}$, $C_{34}$, each with capacitance $C_0$.
$C_{12}$ and $C_{34}$ are in parallel, giving $C_P = C_0 + C_0 = 2C_0$.
This is in series with $C_{23}$, so: \[ C_{eq} = \frac{C_P \cdot C_0}{C_P + C_0} = \frac{2C_0 \cdot C_0}{2C_0 + C_0} = \frac{2C_0^2}{3C_0} = \frac{2}{3}C_0 \] Substituting $C_0 = \dfrac{\varepsilon_0 A}{d}$, we get: \[ C_{eq} = \frac{2\varepsilon_0 A}{3d} \]

Answer 2

Step 1: Understand the problem
We have a configuration of 4 plates forming three capacitors: \(C_{12}\), \(C_{23}\), and \(C_{34}\). Each capacitor has the same capacitance \(C_0 = \frac{\varepsilon_0 A}{d}\).

Step 2: Identify how the plates are connected
- Plates 1 and 3 are connected to terminal \(a\).
- Plates 2 and 4 are connected to terminal \(b\).
This means:
- Capacitors \(C_{12}\) and \(C_{34}\) are effectively in parallel.
- The capacitor \(C_{23}\) is in series with this parallel combination.

Step 3: Calculate equivalent capacitance of the parallel capacitors \(C_{12}\) and \(C_{34}\)
Since capacitors in parallel add directly:
\[ C_P = C_{12} + C_{34} = C_0 + C_0 = 2C_0 \]

Step 4: Calculate equivalent capacitance of \(C_P\) in series with \(C_{23}\)
For capacitors in series, the equivalent capacitance \(C_{eq}\) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_P} + \frac{1}{C_{23}} = \frac{1}{2C_0} + \frac{1}{C_0} = \frac{1}{2C_0} + \frac{2}{2C_0} = \frac{3}{2C_0} \]
Therefore:
\[ C_{eq} = \frac{2C_0}{3} \]

Step 5: Substitute the value of \(C_0\)
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
So:
\[ C_{eq} = \frac{2}{3} \times \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{3 d} \]

Final Result:
\[ \boxed{C_{eq} = \frac{2 \varepsilon_0 A}{3 d}} \]