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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

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

A 5.80-\(\mu\)F, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m\(^3\).

A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

A 20.0-\(\mu\)F capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-\(\mu\)F capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

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