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Capacitors \(C_{1}=6.00 \mu \mathrm{F}\) and \(C_{2}=2.00 \mu \mathrm{F}\) are charged as a parallel combination across a \(250-\mathrm{V}\) battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

Short Answer

Expert verified
The resulting charge on capacitor \( C_{1} \) is 1.5 mC, and on capacitor \( C_{2} \) is 0.5 mC.

Step by step solution

01

Calculate the original charge on each capacitor

The charge on a capacitor is given by the formula: \( Q = C \times V \), where \( Q \) is the charge, \( C \) is the capacitance, and \( V \) is the voltage. Since both capacitors are charged across the same voltage, we can calculate their individual charges. \( Q_{1} = C_{1} \times V = 6.00 \times 10^{-6} F \times 250 V = 1.5 \times 10^{-3} C \), and \( Q_{2} = C_{2} \times V = 2.00 \times 10^{-6} F \times 250 V = 0.5 \times 10^{-3} C \).
02

Calculate the total charge

The total charge is the sum of the charges on both capacitors because they were connected in parallel initially. \( Q_{\text{total}} = Q_{1} + Q_{2} = 1.5 \times 10^{-3} C + 0.5 \times 10^{-3} C = 2.0 \times 10^{-3} C \).
03

Find the resulting charge on each capacitor

When capacitors are connected positive plate to negative plate and vice versa, the total charge is conserved. Since the capacitances are different, the voltage across each capacitor will adjust until charge conservation and electrostatic equilibrium are reached. The resulting charge on each capacitor is the same as the total charge, but split based on the inverse ratio of their capacitances (since now they are effectively in series). The charge will distribute such that \( V_{1} = V_{2} \) and \( Q_{1} = C_{1} \times V_{1} \), \( Q_{2} = C_{2} \times V_{2} \) with the constraint that \( Q_{1} + Q_{2} = Q_{\text{total}} \ \) Rewriting the voltage as a function of charge and capacitance, we get \( Q_{1}/C_{1} = Q_{2}/C_{2} \) and using charge conservation, we have \( Q_{1} + Q_{2} = 2.0 \times 10^{-3} C \). From \( Q_{1}/C_{1} = Q_{2}/C_{2} \), we get \( Q_{1} = Q_{2} \times (C_{1}/C_{2}) \). Substituting this in the charge conservation equation we get \( Q_{2} \times (C_{1}/C_{2}) + Q_{2} = 2.0 \times 10^{-3} C \), or \( Q_{2} = \frac{2.0 \times 10^{-3} C}{(C_{1}/C_{2}) + 1} \).
04

Calculate the final charge on each capacitor

Substituting the capacitances into the formula from Step 3: \( Q_{2} = \frac{2.0 \times 10^{-3} C}{(6.00/2.00) + 1} = \frac{2.0 \times 10^{-3} C}{3 + 1} = 0.5 \times 10^{-3} C \). Now using the charge on \( C_{2} \) and the charge conservation, we can find \( Q_{1} \): \( Q_{1} = 2.0 \times 10^{-3} C - 0.5 \times 10^{-3} C = 1.5 \times 10^{-3} C \). Hence, the resulting charge on \( C_{1} \) and \( C_{2} \) are 1.5 mC and 0.5 mC, respectively.

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Key Concepts

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

Capacitance
Capacitance, in simple terms, is the ability of a system to store an electrical charge. It's a fundamental concept for understanding electronic and electrical systems, particularly when dealing with capacitors. A capacitor is composed of two conductive plates separated by an insulating material, known as the dielectric. When a voltage is applied across these plates, an electric field is established, causing an accumulation of positive charge on one plate and a negative charge on the other.

The capacitance of a capacitor is defined by the equation: \( C = \frac{Q}{V} \), where \( C \) is the capacitance in farads (F), \( Q \) is the charge stored in coulombs (C), and \( V \) is the voltage across the capacitor in volts (V). The capacitance essentially depends on the surface area of the plates, the distance between them, and the properties of the dielectric material. Increased capacitance allows for more charge to be stored at a given voltage. This is a key factor when calculating charges in capacitors, as shown in the exercise problem you're working on.
Parallel and Series Circuits
Understanding how capacitors are connected in circuits is crucial because it determines how they store and distribute charge. Capacitors can be connected in two principal ways: in parallel or in series.

In Parallel

When capacitors are connected in parallel, their total capacitance is the sum of the individual capacitances. Moreover, they all experience the same voltage across their terminals. The charge stored in each capacitor can be different, correlating to their individual capacitances. The total charge in the parallel circuit is thus the sum of the charges on each capacitor, as \( Q_{\text{total}} = Q_{1} + Q_{2} + ... + Q_{n} \). This is reflected in the exercise problem where the initial charge on each capacitor was calculated and then summed to determine the total charge.

In Series

When capacitors are connected in series, the situation becomes a bit more complex. The total capacitance of a series circuit is less than that of the smallest individual capacitor and is calculated based on the reciprocal of the sum of the reciprocals of the individual capacitances. The voltage divides across the capacitors inversely proportional to their capacitance, but interestingly, the charge on all capacitors in a series connection is the same. This is pivotal in the exercise problem where after rearrangement, the capacitors are in series, and the final identical charge is derived based on their respective capacitances.
Charge Conservation
The principle of charge conservation is a foundational concept in physics and electrical engineering. It states that electrical charge can neither be created nor destroyed; it can only be transferred from one place to another. This law is as fundamental as the conservation of energy and is important for analyzing electrical circuits like the one in our exercise.

When the two capacitors from the problem are disconnected and then reconnected in an opposite configuration, the total charge is conserved. No charge is lost in the process; it is simply redistributed between the capacitors. Since the capacitors have different capacitances and are connected in series, charge conservation dictates that the sum of the charges on the individual capacitors after reconnection must equal the total charge they had when initially charged in parallel.

This is used in the final step of the problem's solution to understand how the charges redistribute after the capacitors are connected in series, ensuring that the total charge remains constant throughout the process. Remember, charge conservation helps ensure the underlying stability in systems and confirms that equations balance out properly when analyzing circuits.

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Most popular questions from this chapter

One conductor of an overhead electric transmission line is a long aluminum wire 2.40 \(\mathrm{cm}\) in radius. Suppose that at a particular moment it carries charge per length 1.40\(\mu \mathrm{C} / \mathrm{m}\) and is at potential 345 \(\mathrm{kV}\) . Find the potential 12.0 \(\mathrm{m}\) below the wire. Ignore the other conductors of the transmission line and assume the electric field is everywhere purely radial.

To repair a power supply for a stereo amplifier, an electronics technician needs a \(100-\mu \mathrm{F}\) capacitor capable of withstanding a potential difference of 90 \(\mathrm{V}\) between the plates. The only available supply is a box of five \(100-\mu \mathrm{F}\) capacitors, each having a maximum voltage capability of 50 V. Can the technician substitute a combination of these capacitors that has the proper electrical characteristics? If so, what will be the maximum voltage across any of the capacitors used? (Suggestion: The technician may not have to use all the capacitors in the box.)

Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of 10.0 \(\mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C}\) ?

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrilator (Fig. 26.14\()\) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about 8 \(\mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of 300 \(\mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?

A parallel-plate capacitor of plate separation \(d\) is charged to a potential difference \(\Delta V_{0} .\) A dielectric slab of thickness \(d\) and diectric constant \(\kappa\) is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is \(U / U_{0}=\kappa\) Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Note that this situation is not the same as in Example 26.7 , in which the battery was removed from the circuit before the dielectric was introduced.)

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