Animated Solution for Physics - Electrostatics Potential and Capacitance: Four identical rectangular plates with length, $l=2$ cm and breadth, $b=\frac{3}{2}$ cm are arranged as shown in figure. The equivalent capacitance between A and C is $\frac{x{\epsilon }_0}{d}$. The value of $x$ is ......... . (Round off to the nearest integer)

A Step-by-Step Guide

Multi-plate capacitor problems often look intimidating at first glance. You see a stack of parallel plates, some mysterious connecting wires, and suddenly, finding the equivalent capacitance feels like untangling a complex web. But don't worry! By breaking the physical setup down into a simple electrical circuit, these problems become incredibly straightforward.

Let's dive into this specific problem and see how a systematic approach makes it easy.

Imagine you are looking at the four identical rectangular plates arranged parallel to each other. The first thing to realize is that every pair of adjacent plates forms a distinct capacitor.

Since we have four plates, we have exactly three gaps between them. This means we are dealing with three individual capacitors, which we can call $C_1$, $C_2$, and $C_3$.

Because all the plates are identical (with length $l$ and breadth $b$) and the distance $d$ between any two adjacent plates is the same, each of these three capacitors has the exact same capacitance. Let's call this base capacitance $C_0$:

This is where the magic happens. To convert the physical plates into a standard circuit diagram, we need to trace the electrical nodes.

- Node A: The first plate (Plate A) is our starting terminal. - Node C: The third plate (Plate C) is our ending terminal. - The Catch: Look closely at the diagram. There is a conducting wire connecting the top of Plate B to the top of Plate D. In the world of circuits, a perfect wire means the connected components are at the exact same electrical potential. Therefore, Plate B and Plate D form a single electrical node.

Now, let's redraw the circuit based on our nodes:

1. Capacitor $C_1$ is physically between Plate A and Plate B. So, it connects Node A to Node B. 2. Capacitor $C_2$ is physically between Plate B and Plate C. So, it connects Node B to Node C. 3. Capacitor $C_3$ is physically between Plate C and Plate D. But wait! We established that Plate D is electrically the same as Plate B. Therefore, $C_3$ is effectively connected between Node C and Node B.

Do you see what just happened? Both $C_2$ and $C_3$ are connected across the exact same two nodes (B and C). This is the textbook definition of a parallel combination!

First, let's resolve the parallel part. For capacitors in parallel, we simply add their capacitances:

Now, our circuit is simplified to just two components in series: $C_1$ (which is $C_0$) and our new equivalent capacitor $C_{23}$ (which is $2C_0$). We use the series formula to find the total equivalent capacitance $C_{eq}$:

Flipping this over gives us:

We are almost there! We just need to substitute the actual value of $C_0$ back into our equation. First, let's calculate the area $A$ of the plates:

Now, substitute this into our $C_{eq}$ expression:

The $3$ in the numerator and the $3$ in the denominator cancel out perfectly, leaving us with:

The problem states that the equivalent capacitance is $\frac{x{\epsilon }_0}{d}$. By comparing our result with this expression, it is clear that $x=2$.

And there you have it! By carefully labeling nodes and redrawing the circuit, a confusing stack of plates turns into a simple series-parallel problem.