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Ch 26: Potential and Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 26: Potential and FieldProblem 63
Chapter 26, Problem 63

Capacitors C₁ = 10 μF and C₂ = 20 μF are each charged to 10 V, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C₁ connected to the negative plate of C₂ and vice versa. Afterward, what are the charge on and the potential difference across each capacitor?

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Understand the initial conditions: Capacitor C₁ has a capacitance of 10 μF and is charged to a voltage of 10 V, so its initial charge is given by Q₁ = C₁ × V₁. Similarly, capacitor C₂ has a capacitance of 20 μF and is charged to a voltage of 10 V, so its initial charge is Q₂ = C₂ × V₂. Note that the charges on the capacitors are opposite in polarity when connected as described in the problem.
Determine the total charge in the system after connection: When the capacitors are connected in parallel with opposite polarities, the total charge in the system is conserved. The net charge is Q_total = Q₁ - Q₂ because the charges are opposite in polarity.
Calculate the equivalent capacitance of the parallel combination: When capacitors are connected in parallel, their equivalent capacitance is the sum of their individual capacitances. Thus, C_eq = C₁ + C₂.
Find the final voltage across the capacitors: After the capacitors are connected, they will share the same voltage because they are in parallel. The final voltage V_f can be found using the relationship Q_total = C_eq × V_f, where Q_total is the net charge calculated earlier.
Determine the final charge on each capacitor: Once the final voltage V_f is known, the charge on each capacitor can be calculated using the formula Q = C × V. For C₁, the final charge is Q₁_f = C₁ × V_f, and for C₂, the final charge is Q₂_f = C₂ × V_f.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage, measured in farads (F). It is defined by the formula C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the capacitor. In this scenario, the capacitance values of C₁ and C₂ are given as 10 μF and 20 μF, respectively, which will influence the charge distribution when they are connected.
Recommended video:
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08:02
Capacitors & Capacitance (Intro)

Charge Conservation

Charge conservation is a fundamental principle stating that the total electric charge in an isolated system remains constant. When the two capacitors are connected in parallel with opposite plates, the charges will redistribute between them while maintaining the total charge. This principle is crucial for determining the final charge on each capacitor after they are connected.
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Conservation of Charge

Voltage in Parallel Circuits

In a parallel circuit, the voltage across each component is the same. When capacitors are connected in parallel, the potential difference across each capacitor will equal the voltage across the entire arrangement. This concept is essential for calculating the final voltage across each capacitor after they have been connected, as it will help determine how the initial charges affect the final state.
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Resistors in Parallel in an AC Circuit
Related Practice
Textbook Question

What are the charge on and the potential difference across each capacitor in FIGURE P26.57?

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Textbook Question

Initially, the switch in FIGURE P26.61 is in position A and capacitors C₂ and C₃ are uncharged. Then the switch is flipped to position B. Afterward, the voltage across C₁ is 4.0 V. What is the emf of the battery?

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Textbook Question

Six identical capacitors with capacitance C are connected as shown in FIGURE P26.59. What is the potential difference between points a and b?

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Textbook Question

An isolated 5.0 μF parallel-plate capacitor has 4.0 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.0 μF. How much work is done by the external force?

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Textbook Question

You've built a device that uses the energy from a rapidly discharged capacitor to launch the capacitor straight up. One capacitor, with a mass of 3.5 g, is launched to a height of 1.6 m after having been charged to 100 V. What is its capacitance in μF?

Textbook Question

The label rubbed off one of the capacitors you are using to build a circuit. To find out its capacitance, you place it in series with a 10 μF capacitor and connect them to a 9.0 V battery. Using your voltmeter, you measure 6.0 V across the unknown capacitor. What is the unknown capacitor's capacitance?

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