Animated Solution for Physics - Electrostatics: Five identical capacitor plates, each of area $A$, are arranged such that adjacent plates are at a distance $d$ apart, the plates are connected to a source of emf $V$ as shown in the figure. The charge on plate 1 is ...... and on plate 4 is .......

A Stack of Capacitors

Imagine you have a loaf of bread, and you take five slices and stand them up parallel to each other. In our physics problem, these slices are identical conducting plates, each with an area $A$, separated by a small distance $d$.

When you place five plates next to each other, the spaces between them form the actual capacitors. Between plate 1 and 2 is the first capacitor, between 2 and 3 is the second, and so on. Therefore, five plates create exactly four individual parallel plate capacitors.

The capacitance of each of these individual capacitors is given by the standard formula:

Now, let's trace the wires. Plates 1, 3, and 5 are all wired together and connected to one terminal of the battery. Plates 2 and 4 are wired together and connected to the other terminal.

What does this mean physically? It means that every single capacitor has one plate connected to the positive terminal and its other plate connected to the negative terminal. This is the classic definition of a parallel combination. Because they are in parallel, the potential difference across every single capacitor is exactly $V$.

Since the potential difference is $V$ and the capacitance is $C$, the magnitude of the charge on each face of any capacitor is:

Analyzing Plate 1: Plate 1 is an outermost plate. It only has one face that looks at another plate (Plate 2). The outer face of Plate 1 doesn't form a capacitor. Since Plate 1 is connected to the positive terminal, its inner face will acquire a positive charge. Therefore, the total charge on Plate 1 is simply $+q$, which equals $\frac{{\epsilon }_{0}AV}{d}$.

Analyzing Plate 4: Plate 4 is an inner plate. It is sandwiched between Plate 3 and Plate 5. This means both of its faces are actively participating in capacitors. Since Plate 4 is connected to the negative terminal, the face looking at Plate 3 gets a charge of $-q$, and the face looking at Plate 5 also gets a charge of $-q$. Therefore, the total charge on Plate 4 is the sum of the charges on its two faces:

By carefully analyzing the faces of the plates and their connections to the battery, we can confidently state the final charges:

Charge on Plate 1: $\frac{{\epsilon }_{0}AV}{d}$ Charge on Plate 4: $-\frac{2{\epsilon }_{0}AV}{d}$