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You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

Short Answer

Expert verified
In parallel, both energy and charge are four times that in series. The electric field ratio series:parallel is 1:2.

Step by step solution

01

Understand the Basic Capacitor Formulas

The energy stored in a capacitor is given by the formula \(U = \frac{1}{2}CV^2\). Capacitors store charge given by \(Q = CV\). When capacitors are in series and in parallel, their overall capacitance value changes accordingly.
02

Series Capacitor Configuration Calculation

For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals of each individual capacitance: \(\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}\). Therefore, \(C_s = \frac{C}{2}\). The total energy stored in the capacitors when connected in series is \(U_s = \frac{1}{2}C_sV^2 = \frac{1}{2}\cdot\frac{C}{2}\cdot V^2 = \frac{CV^2}{4}\). The charge stored is \(Q_s = C_sV = \frac{CV}{2}\).
03

Parallel Capacitor Configuration Calculation

For capacitors in parallel, the total capacitance is the sum of the individual capacitances: \(C_p = C + C = 2C\). The total energy stored in the capacitors when connected in parallel is \(U_p = \frac{1}{2}C_pV^2 = \frac{1}{2}\cdot 2C \cdot V^2 = CV^2\). The charge stored is \(Q_p = C_pV = 2CV\).
04

Compare Energies of Series and Parallel

In series, the energy stored is \(\frac{CV^2}{4}\), while in parallel it is \(CV^2\). Thus, energy stored in parallel is four times that in series.
05

Compare Charges of Series and Parallel

In series, the charge stored is \(\frac{CV}{2}\), and in parallel, it is \(2CV\). Therefore, the charge stored in parallel is four times that stored in series.
06

Electric Field Ratio Calculation

The electric field \(E\) in a capacitor is given by \(E = \frac{V}{d}\), where \(d\) is the separation between the plates. In the series combination, the voltage is divided across the capacitors, whereas in parallel, each gets the full voltage \(V\). Thus, for two capacitors, the ratio of the electric field for series to parallel is \(\frac{V/2d}{V/d} = \frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitors in Series
When capacitors are connected in series, they share the same charge but the total voltage across them is the sum of voltages across each capacitor. This configuration affects their overall capacitance significantly.
The formula for total capacitance in series is given by the inverse of the sum of the inverses of each individual capacitance:
  • \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + ... \)
This means for two identical capacitors, each with capacitance \(C\) in series:
  • \( C_s = \frac{C}{2} \)
The stored energy in this series configuration is quite interesting as it's less efficient compared to when the same capacitors are in parallel. The energy stored can be derived as:
  • \( U_s = \frac{CV^2}{4} \)
This shows that storing energy with capacitors in series might require careful consideration depending on the application. This is because the energy storage efficiency is less than their parallel counterpart."The charge stored in each capacitor in series is also different from their parallel configuration:
  • \( Q_s = \frac{CV}{2} \)
Capacitors in Parallel
Connecting capacitors in parallel leads them to have the same voltage across each capacitor while the charge is shared among them, boosting the system's total capacitance.
For capacitors in parallel, the effective capacitance is simply the sum of all individual capacitances:
  • \( C_p = C_1 + C_2 + ... \)
For two identical capacitors, each with capacitance \(C\), their total parallel capacitance is:
  • \( C_p = 2C \)
More importantly, the energy storage capabilities of capacitors in parallel are significantly enhanced compared to those in series. The energy stored in this configuration is:
  • \( U_p = CV^2 \)
This indicates that the parallel configuration allows the capacitors to store more energy, making it practical for applications needing high energy storage. Additionally, the total charge stored here is:
  • \( Q_p = 2CV \)
These characteristics of parallel capacitors underscore their utility in circuits that require robust energy storage solutions.
Electric Field in Capacitors
The electric field within a capacitor is a crucial factor that dictates its overall voltage and energy characteristics. This electric field \(E\) is determined by the voltage \(V\) across the capacitor plates and the separation distance \(d\) between them:
  • \( E = \frac{V}{d} \)
In a series configuration, because the capacitors share the overall voltage, the voltage effectively across each individual capacitor is divided, leading to an electric field half of what it would be in parallel:
  • Series Electric Field: \( E_s = \frac{V/2}{d} \)
  • Parallel Electric Field: \( E_p = \frac{V}{d} \)
This results in an electric field ratio for series to parallel configurations of:
  • \( \text{Ratio} = \frac{1}{2} \)
This ratio is crucial when considering applications where the electric field's strength is a limiting factor, especially in high-voltage applications. Understanding how configuration affects the electric field can help in designing circuits that are both safe and efficient.

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Most popular questions from this chapter

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

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