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

Knight Calc 5th Edition

Ch 26: Potential and Field

Problem 67

Chapter 26, Problem 67

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?

Verified step by step guidance

Step 1: Recall the formula for capacitors in series. The total capacitance of capacitors in series is given by:

\frac{1}{C} = \frac{1}{C}

₁+1C₂, where

C

₁ and

C

₂ are the capacitances of the individual capacitors.

Step 2: Understand that the voltage across capacitors in series is divided based on their capacitances. The voltage across a capacitor is inversely proportional to its capacitance:

V

₁ =

\frac{Q}{C}

₁ and

V

₂ =

\frac{Q}{C}

₂, where

Q

is the charge on the capacitors.

Step 3: Use the given information: the total voltage across the series combination is 9.0 V, and the voltage across the unknown capacitor is 6.0 V. This means the voltage across the 10 μF capacitor is

9.0 - 6.0 = 3.0

Step 4: Apply the relationship between voltage and capacitance for capacitors in series. Since the charge

Q

is the same for both capacitors, the ratio of voltages is inversely proportional to the ratio of capacitances:

V

₁V₂ =

C

₂C₁. Substitute

V

₁ = 3.0 V,

V

₂ = 6.0 V, and

C

₁ = 10 μF.

Step 5: Solve for the unknown capacitance

C

₂ using the ratio derived in Step 4. Then, verify the total capacitance of the series combination using the formula from Step 1 to ensure consistency.

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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 electrical charge per unit voltage. It is measured in farads (F) and is defined as the ratio of the charge (Q) stored on one plate of the capacitor to the voltage (V) across the plates, expressed as C = Q/V. Understanding capacitance is crucial for analyzing how capacitors behave in circuits.

Series Capacitors

When capacitors are connected in series, the total capacitance (C_total) is less than the smallest individual capacitance. The formula for total capacitance in series is given by 1/C_total = 1/C1 + 1/C2 + ... + 1/Cn. This concept is essential for determining the overall capacitance when combining capacitors in a circuit.

Voltage Division in Capacitors

In a series circuit, the voltage across each capacitor is divided based on their capacitance values. The voltage across a capacitor in series can be calculated using the formula V = Q/C, where Q is the charge (same for all capacitors in series) and C is the capacitance. This principle helps in understanding how the applied voltage is distributed among the capacitors.

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Related Practice

Textbook Question

High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE P26.68. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius R₁ while the outer conductor of radius R₂ is grounded (i.e., at V=0 V). An insulating material fills the space between them, and an insulating plastic coating goes around the outside. Evaluate the capacitance per meter of a cable having R₁=0.50 mm and R₂=3.0 mm.

Textbook Question

A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C₀. What is the capacitance if an insulator with dielectric constant κ and thickness d/2 is slipped between the electrodes without changing the plate separation?

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

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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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 current that charges a capacitor transfers energy that is stored in the capacitor's electric field. Consider a 2.0 μF capacitor, initially uncharged, that is storing energy at a constant 200 W rate. What is the capacitor voltage 2.0 μs after charging begins?

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