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24.37. 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 inaximuni electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

Short Answer

Expert verified
(a) Energy is higher in parallel by 4 times; (b) Maximum charge is higher in parallel by 4 times; (c) Electric field is smaller in series by half.

Step by step solution

01

Understand Capacitor Combinations

For a parallel combination, the total capacitance is the sum of the individual capacitances: \( C_{parallel} = C_1 + C_2 = 2C \). For a series combination, the total capacitance is given by \( \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2C} \), yielding \( C_{series} = \frac{C}{2} \).
02

Calculate Total Energy Stored in Capacitors

The energy stored in a capacitor is given by \( U = \frac{1}{2} C V^2 \). For parallel connection: \( U_{parallel} = \frac{1}{2} (2C) V^2 = CV^2 \). For series connection: \( U_{series} = \frac{1}{2} \left(\frac{C}{2}\right) V^2 = \frac{CV^2}{4} \). Thus, the energy in parallel is four times the energy in series.
03

Determine Maximum Charge Stored

The charge stored on a capacitor is given by \( Q = CV \). For parallel connection: \( Q_{parallel} = (2C)V = 2CV \). For series connection: \( Q_{series} = \left(\frac{C}{2}\right) V = \frac{CV}{2} \). Thus, the charge in parallel is four times the charge in series.
04

Calculate Electric Field Ratio

The electric field between the plates is given by \( E = \frac{V}{d} \), where \(d\) is the distance between plates. Each capacitor in series has half the total voltage \( \left(\frac{V}{2}\right) \), whereas in parallel each sees the full \( V \). Therefore, \( E_{series} = \frac{V/2}{d} = \frac{V}{2d} \) and \( E_{parallel} = \frac{V}{d} \). The ratio \( \frac{E_{series}}{E_{parallel}} = \frac{1}{2} \).

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Key Concepts

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

Energy Storage in Capacitors
Understanding how energy is stored in capacitors, especially when combined in series and parallel, is key. Capacitors store energy by maintaining an electric field between their plates when a voltage is applied. The energy stored in a capacitor can be calculated using the formula: \[U = \frac{1}{2} C V^2\]Here, \( U \) is the energy stored, \( C \) is the capacitance of the capacitor, and \( V \) is the voltage across it.- **In Parallel:** When capacitors are connected in parallel, their total capacitance increases, resulting in greater energy storage. For our two capacitors, the energy stored becomes: \[ U_{parallel} = CV^2 \]- **In Series:** Conversely, when in series, the effective capacitance is reduced, thus lowering the total energy stored. It is calculated as: \[ U_{series} = \frac{CV^2}{4} \]The series connection stores only a quarter of the energy compared to the parallel connection given the same voltage.
Capacitor Charge Storage
The charge stored in a capacitor is another fundamental concept to grasp. Charge, denoted by \( Q \), depends directly on the capacitor's capacitance and the applied voltage through the following equation:\[Q = CV\]- **Parallel Connection:** For capacitors in parallel, the total capacitance is the sum of both, doubling their charge storage capability. \[ Q_{parallel} = 2CV \] This greater charge storage benefits circuits that need to maintain a higher charge over time.- **Series Connection:** Capacitors in series halve the charge since the effective capacitance is reduced: \[ Q_{series} = \frac{CV}{2} \]In practical applications, choosing between series and parallel arrangements depends on whether maximizing charge or conserving space is priority.
Electric Field in Capacitors
The electric field between capacitor plates is what stores energy in a capacitor. This field is calculated by the voltage between the plates divided by the distance separating them:\[E = \frac{V}{d}\]Where \( E \) is the electric field strength, \( V \) is the voltage, and \( d \) is the distance between plates.- **Series Connection:** In a series setup, the voltage is divided across the capacitors, halving the electric field produced by each. \[ E_{series} = \frac{V}{2d} \]- **Parallel Connection:** In parallel, each capacitor is subjected to the full voltage, maximizing the electric field: \[ E_{parallel} = \frac{V}{d} \]The electric field in series configurations is weaker but can be suitable for sensitive electronic components. Understanding these differences is vital for effective and safe circuit designs.

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Most popular questions from this chapter

24.40. A budding electronics hobbyist wants to make a simple 1.0 -nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few shects of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(120 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 \(\mathrm{nF}\) of capacitance? (c) Suppose she goes hight-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Tellon than of posterboard? Explain.

24.26. An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (d) When the charge is 0.0180\(\mu \mathrm{C}\) , what total energy is stored?

24.61. Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) potential difference. (a) What is the charge on the \(4.2-\mu \mathrm{F}\) capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

24.28. A capacitor of capacitance \(C\) is charged to a potential difference \(V_{0}\) . The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance \(C / 2\) Compute (a) the original charge of the system; (b) the final potential difference across each capacitor; (e) the final energy of the system; (d) the decrease in energy when the capacitors are connected. (e) Where did the "lost" energy go?

24.11. A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

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