Ch 24: Capacitance and Dielectrics

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 24: Capacitance and Dielectrics

Problem 3

Chapter 24, Problem 3

A parallel-plate air capacitor of capacitance $245$ pF has a charge of magnitude $0.148$ $μ$ C on each plate. The plates are $0.328$ mm apart.
(a) What is the potential difference between the plates?
(b) What is the area of each plate?
(c) What is the electric field magnitude between the plates?
(d) What is the surface charge density on each plate?

Verified step by step guidance

To find the potential difference (V) between the plates, use the formula for capacitance: $ C = \frac{Q}{V} $. Rearrange this to solve for V: $ V = \frac{Q}{C} $. Substitute the given values: $ Q = 0.148 \times 10^{-6} \text{ C} $ and $ C = 245 \times 10^{-12} \text{ F} $.

To find the area (A) of each plate, use the formula for the capacitance of a parallel-plate capacitor: $ C = \frac{\varepsilon_0 A}{d} $, where $ \varepsilon_0 $ is the permittivity of free space $ (8.85 \times 10^{-12} \text{ F/m}) $ and $ d = 0.328 \times 10^{-3} \text{ m} $. Rearrange to solve for A: $ A = \frac{C \cdot d}{\varepsilon_0} $.

To find the electric field magnitude (E) between the plates, use the relationship $ E = \frac{V}{d} $. Use the potential difference found in step 1 and the distance $ d = 0.328 \times 10^{-3} \text{ m} $.

To find the surface charge density ($ \sigma $) on each plate, use the formula $ \sigma = \frac{Q}{A} $. Use the charge $ Q = 0.148 \times 10^{-6} \text{ C} $ and the area found in step 2.

Review the relationships and ensure all units are consistent throughout the calculations. This will help verify the correctness of each step and the final results.

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Key Concepts

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

Capacitance and Potential Difference

Capacitance (C) is the ability of a capacitor to store charge per unit voltage, defined by the formula C = Q/V, where Q is the charge and V is the potential difference. To find the potential difference between the plates, rearrange the formula to V = Q/C, using the given values for charge and capacitance.

Electric Field in a Capacitor

The electric field (E) between the plates of a parallel-plate capacitor is uniform and can be calculated using E = V/d, where V is the potential difference and d is the separation between the plates. This relationship shows how the field strength is directly proportional to the voltage and inversely proportional to the distance between the plates.

Surface Charge Density

Surface charge density (σ) is the amount of charge per unit area on a plate, given by σ = Q/A, where Q is the charge and A is the area of the plate. This concept is crucial for understanding how charge is distributed across the surface of the capacitor plates, affecting the electric field and potential difference.

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Surface Charge Density

Related Practice

Textbook Question

A $5.00$ - $μ$ F parallel-plate capacitor is connected to a $12.0$ V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates.

(a) A voltmeter is connected across the two plates without discharging them. What does it read?

(b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

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Textbook Question

A parallel-plate air capacitor is to store charge of magnitude $240.0$ pC on each plate when the potential difference between the plates is $42.0$ V.

(a) If the area of each plate is $6.80$ cm², what is the separation between the plates?

(b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude $240.0$ pC on each plate?

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Textbook Question

The plates of a parallel-plate capacitor are $2.50$ mm apart, and each carries a charge of magnitude $80.0$ nC. The plates are in vacuum. The electric field between the plates has a magnitude of $4.00 \times 10^{6}$ V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

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Textbook Question

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is $10.0$ pC. The inner cylinder has radius $0.50$ mm, the outer one has radius $5.00$ mm, and the length of each cylinder is $18.0$ cm.

(a) What is the capacitance?

(b) What applied potential difference is necessary to produce these charges on the cylinders?

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