Question

Find the equivalent capacitance between A and B. (A : Area of each plate, d : separation between adjacent plates)

Answer 1

3 \frac{\epsilon_0 A}{d}

Answer 2

To find the equivalent capacitance between points A and B, we first identify the potential of each plate based on the connections. Let the plates be numbered 1 to 5 from top to bottom. The capacitance between any two adjacent plates is $C_0 = \frac{\epsilon_0 A}{d}$.

1. Identify the potentials of the plates:

   • Plate 1 is connected to terminal A. So, its potential is $V_A$.
   • Plate 3 is connected to terminal A. So, its potential is $V_A$.
   • Plate 5 is connected to terminal B. So, its potential is $V_B$.
   • Plate 2 is connected to a common node on the left side. Let's call its potential $V_X$.
   • Plate 4 is connected to the same common node on the left side (potential $V_X$) AND it is also directly connected to terminal B. This means that the common node $X$ is at the same potential as B. Therefore, $V_X = V_B$.

2. Determine the potential of each plate:

   • Plate 1: $V_A$
   • Plate 2: $V_X = V_B$
   • Plate 3: $V_A$
   • Plate 4: $V_X = V_B$
   • Plate 5: $V_B$

3. Identify the individual capacitors and their connections between A and B:

   • Capacitor between Plate 1 and Plate 2 ($C_{12}$): This capacitor is between Plate 1 (at $V_A$) and Plate 2 (at $V_B$). So, $C_{12}$ is connected between A and B. Its capacitance is $C_0$.
   • Capacitor between Plate 2 and Plate 3 ($C_{23}$): This capacitor is between Plate 2 (at $V_B$) and Plate 3 (at $V_A$). So, $C_{23}$ is connected between B and A (which is the same as between A and B). Its capacitance is $C_0$.
   • Capacitor between Plate 3 and Plate 4 ($C_{34}$): This capacitor is between Plate 3 (at $V_A$) and Plate 4 (at $V_B$). So, $C_{34}$ is connected between A and B. Its capacitance is $C_0$.
   • Capacitor between Plate 4 and Plate 5 ($C_{45}$): This capacitor is between Plate 4 (at $V_B$) and Plate 5 (at $V_B$). Since both plates are at the same potential ($V_B$), there is no potential difference across this capacitor. Thus, it does not store any charge and effectively does not contribute to the equivalent capacitance between A and B. It is effectively short-circuited.

4. Calculate the equivalent capacitance: We have three capacitors ($C_{12}$, $C_{23}$, and $C_{34}$) each of capacitance $C_0$ connected in parallel between points A and B. For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances: $C_{eq} = C_{12} + C_{23} + C_{34}$ $C_{eq} = C_0 + C_0 + C_0$ $C_{eq} = 3 C_0$

   Substituting the value of $C_0 = \frac{\epsilon_0 A}{d}$: $C_{eq} = 3 \frac{\epsilon_0 A}{d}$

The final answer is $3 \frac{\epsilon_0 A}{d}$.

Explanation of the solution: The five plates form four parallel plate capacitors. By tracing the connections, plates 1 and 3 are connected to A. Plates 2 and 4 are connected to a common node on the left, and plate 4 is also connected to B. This implies the common node on the left is at the same potential as B. Thus, the capacitors formed by (plate 1, plate 2), (plate 2, plate 3), and (plate 3, plate 4) are all connected between A and B in parallel. The capacitor formed by (plate 4, plate 5) is connected between B and B, hence it is shorted and does not contribute. Each active capacitor has capacitance $C_0 = \frac{\epsilon_0 A}{d}$. Since they are in parallel, the equivalent capacitance is $3C_0 = 3 \frac{\epsilon_0 A}{d}$.