Question

Fifty parallel plate capacitors are connected in series. The distance between the plates is \(d\) for the first capacitor, \(2 d\) for the second capacitor, \(3 d\) for the third capacitor, and so on. The area of the plates is the same for all the capacitors. Express the equivalent capacitance of the whole set in terms of \(C_{1}\) (the capacitance of the first capacitor).

Short answer

Answer: The equivalent capacitance is given by the expression \(C_{eq} = \frac{C_1}{1275}\).

Step-by-step solution

Understanding capacitors connected in series

When multiple capacitors are connected in series, the total capacitance can be calculated using the formula: \[\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + ... + \frac{1}{C_{n}}\] Where: - \(C_{eq}\) is the equivalent capacitance - \(C_{1}, C_{2},..., C_{n}\) are the capacitances of the capacitors connected in series.

Finding the relationship between each capacitor's capacitance

It is given that the area of the plates is the same for all capacitors, and the distance between the plates increases in a linear manner (\({d, 2d, 3d,\dots}\)). So, the capacitance of a capacitor can be defined using the formula: \[C = \frac{A \epsilon_0}{d}\] Where: - \(C\) is the capacitance - \(A\) is the area of the plates - \(\epsilon_0\) is the vacuum permittivity - \(d\) is the distance between the plates Since the area and vacuum permittivity are constant for each capacitor, we can observe that the capacitance is inversely proportional to the distance between the plates. So, for the \(i^{th}\) capacitor with a plate distance of \(id\), we can write: \[C_i = \frac{C_1}{i}\]

Calculating the equivalent capacitance

Now, we have the relationship between each capacitor's capacitance and the first capacitor's capacitance. We can plug this relationship into the series capacitance formula from step 1: \[\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{\frac{C_1}{2}} + \frac{1}{\frac{C_1}{3}} + ... + \frac{1}{\frac{C_1}{50}}\] We can write this as a summation: \[\frac{1}{C_{eq}} = \sum_{i=1}^{50} \frac{1}{\frac{C_1}{i}}\] Now, we can simplify this expression by multiplying both sides by \(C_1\): \[C_1 = \sum_{i=1}^{50} i\] This sum can be evaluated using the arithmetic series sum formula: \[C_1 = \frac{n(n+1)}{2}\] Where \(n = 50\): \[C_1 = \frac{50 \cdot (50+1)}{2}\] Now, we can solve for the equivalent capacitance, \(C_{eq}\): \[C_{eq} = \frac{C_1}{1275}\] And finally, we have the expression for the equivalent capacitance of the whole set of capacitors in terms of \(C_{1}\): \[C_{eq} = \frac{C_1}{1275}\]

Key concepts

Series Capacitors

When capacitors are connected in series, their overall behavior can be quite different compared to a single capacitor. In a series configuration, the capacitors share the same charge but have different voltage drops across each capacitor. The total or equivalent capacitance (C_{eq}) of the entire system is less than any single capacitor's capacitance. This is due to how charge distribution across the capacitors impacts the total storage capacity. Imagine this as capacitors aligned end-to-end, where each seems to stretch the system's total capacity. This results in a smaller total capacitance, meaning the system stores less charge for the same voltage compared to a single capacitor of larger value. To find the equivalent capacitance in series, we use the formula:\[\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \ldots + \frac{1}{C_{n}} \]Here, each term on the right side represents the inverse of individual capacitance. If you image-charge these capacitors like buckets, with each additional in series, less total water (or charge) can be held without leaking due to balancing the distributed load.

Capacitance Formula

Capacitance (C) is a measure of how much electric charge a capacitor can store per unit voltage. For parallel plate capacitors, the formula is:\[C = \frac{A \epsilon_0}{d}\]where:

• C stands for capacitance
• A is the area of one of the plates
• \(\epsilon_0\) is the vacuum permittivity, a physical constant
• d is the distance between the plates

Understanding this formula reveals that capacitance depends directly on the plate area and inversely on the distance between them. This means larger areas or smaller distances result in higher capacitance, allowing the capacitor to store more charge at the same voltage. For instance, halving the distance between plates doubles the capacitance, demonstrating the advantage of tightly packed plate arrangements in real-world capacitors.

Arithmetic Series

An arithmetic series is a sequence of numbers where each term after the first is created by adding a constant difference to the previous term. This concept plays a crucial role when calculating the total sum needed for finding equivalent capacitance in our setup. In the exercise, the distance between the capacitor plates increases linearly by factors of d (e.g., 1d, 2d, 3d, ..., 50d). As such, the individual capacitance values are inversely related to these factors, forming a reciprocal arithmetic series when substituted back into the equivalent capacitance formula.To sum up the reciprocals of integers (1 through 50), the arithmetic series formula is applied:\[S = \frac{n(n+1)}{2}\]where n is the number of terms, in this case, 50. This sum becomes essential for calculating the total equivalent capacitance, simplifying the otherwise cumbersome manual addition of each inverse term.

Capacitor Plate Distance

In capacitor design, the distance between the plates shows up as a crucial parameter influencing the capacitance value. In a parallel plate capacitor, shortening the gap between plates usually increases capacitance because more electric field lines are concentrated between the plates, leading to better charge storage capability. In our exercise, this distance becomes a sequence multiplying d by integers (1, 2, 3, ..., up to 50), making each successive capacitor in the series less capable of storing charge. This setup highlights the inverse relationship between capacitance and plate separation; as the distance increases, capacitance diminishes sharply. Knowing this helps in designing capacitors where adjusting plate distance can fine-tune desired electrical characteristics. Simple adjustments can yield profound changes in how effectively a capacitor can hold and maintain charge, which is crucial in both small scale circuits and large industrial applications.