Problem 115 - Physics

Question

A sheet of gold foil (negligible thickness) is placed between the plates of a capacitor and has the same area as each of the plates. The foil is parallel to the plates, at a position one-third of the way from one to the other. Before the foil is inserted, the capacitance is \(C_{0}\). What is the capacitance after the foil is in place? Express your answer in terms of \(C_{0}\) -

Step-by-Step Solution

Answer

The capacitance after inserting the foil is \( \frac{6C_0}{5} \).

Step 1: Understanding the Problem

We have a capacitor with a certain capacitance, denoted as \( C_0 \). A thin sheet of conductive gold foil is introduced into the space between the capacitor plates. This foil divides the original capacitor into two capacitors in series, each having the same plate area as the original plates but different distances between them.

Step 2: Analyzing Capacitor Arrangement

The gold foil divides the original distance \( d \) between the plates into two segments: \( \frac{d}{3} \) and \( \frac{2d}{3} \). This effectively creates two capacitors in series: \( C_1 \) and \( C_2 \), with plate separations \( \frac{d}{3} \) and \( \frac{2d}{3} \), respectively.

Step 3: Expressing Capacitance of Sub-capacitors

For a parallel-plate capacitor, capacitance \( C \) is given by the formula \( C = \frac{\varepsilon A}{d} \). Hence, \( C_1 = \frac{\varepsilon A}{d/3} = 3\frac{\varepsilon A}{d} = 3C_0 \) and \( C_2 = \frac{\varepsilon A}{2d/3} = \frac{3}{2} \frac{\varepsilon A}{d} = \frac{3}{2} C_0 \).

Step 4: Calculating the Equivalent Capacitance of Series Capacitors

The total capacitance \( C \) for capacitors in series is given by: \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \). Substituting for \( C_1 \) and \( C_2 \), we have \( \frac{1}{C} = \frac{1}{3C_0} + \frac{2}{3C_0} \). Combine these to find \( \frac{1}{C} = \frac{5}{6C_0} \).

Step 5: Solving for Total Capacitance

From \( \frac{1}{C} = \frac{5}{6C_0} \), solve for \( C \): \( C = \frac{6C_0}{5} \). This is the effective capacitance of the system after inserting the foil between the plates of the capacitor.

Key Concepts

Parallel-plate capacitor

The parallel-plate capacitor is a fundamental component in electronics, commonly used for storing electrical energy. Imagine two flat plates, facing each other, separated by a small distance. These plates hold equal and opposite charges, creating an electric field between them. This setup is what makes up a parallel-plate capacitor.
To determine how much energy the capacitor can store, we calculate its capacitance \(C\). This is given by the formula \(C = \frac{\varepsilon A}{d}\), where \(\varepsilon\) is the permittivity of the material between the plates, \(A\) is the area of one of the plates, and \(d\) is the separation between them. The larger the plate area and the closer they are, the higher the capacitance.
Capacitors in and of themselves are intriguing because according to their arrangement, they can influence the behavior of the circuit in different ways.

Series capacitors

When capacitors are in series, the arrangement essentially involves stacking them one after the other. This is analogous to the situation where the gold foil divides the original capacitor into two new ones, separated by different distances. Just like dominoes lined up in a row, charge must flow through each component sequentially.
For capacitors in series, the overall or equivalent capacitance is not simply the sum of the individual capacitances. Instead, you use the formula for series capacitance: \[\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\]where \(C\) is the total capacitance, and \(C_1, C_2, \ldots, C_n\) are the individual capacitances. This is because in a series, the same charge passes through all capacitors, leading to a unique way of calculating total capacitance.

Dielectric constant

The dielectric constant characterizes the material inserted between the plates of a capacitor. It measures the material's ability to increase the capacitor's capacitance compared to vacuum. Imagine different materials being like lenses for light; they all bend the light differently, and similarly, dielectrics affect electricity in capacitors.
Dielectrics boost the capacitance because they reduce the electric field across the space between the plates, allowing the plates to store more charge for the same voltage. The dielectric constant \(\varepsilon_{r}\) quantifies this, and the overall capacitance becomes \(C = \varepsilon_{r} \frac{\varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the vacuum permittivity.

Capacitor rearrangement

Capacitor rearrangement refers to the method of altering the geometry or configuration of a capacitor setup to modify its electrical properties, often for a specific purpose like increasing or decreasing capacitance. This concept is beautifully exemplified by inserting a conductive foil into the capacitor, as seen in the exercise.
By inserting the foil, the original capacitor is split into two series capacitors, each with its own distinct distance. This rearrangement effectively changes how the electrical energy is distributed between them and impacts the total capacitance. Understand that such modifications do not always follow a linear path and usually require recalculations of \(C\) to understand the new setup better.