Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 24 - Capacitance, Dielectrics, Electric Energy, Storage

Problem 85b

Chapter 23, Problem 85b

A parallel-plate capacitor with plate area 2.0 cm² and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. The plates are now pulled to a separation of 0.85 mm. What is the charge on the capacitor now?

Verified step by step guidance

Understand the problem: A parallel-plate capacitor is initially charged by a 12-V battery, then disconnected. The plate separation is increased, and we need to determine the charge on the capacitor after the separation change.

Recall that the charge on a capacitor is given by the formula:

Q = C V

, where

Q

is the charge,

C

is the capacitance, and

V

is the voltage. Since the battery is disconnected, the charge

Q

remains constant regardless of the plate separation.

The capacitance of a parallel-plate capacitor is given by:

C = \frac{ε}{d}

, where

ε

is the permittivity of free space (

8.85 \times 10^{- 12}

F/m),

A

is the plate area, and

d

is the plate separation.

Substitute the given values into the capacitance formula to calculate the initial capacitance:

C = \frac{ε}{d}

, where

A = 2.0 \times 10^{- 4}

m² and

d = 0.50 \times 10^{- 3}

Since the charge remains constant after the battery is disconnected, the charge on the capacitor is the same as it was initially. Use the formula

Q = C V

with the initial capacitance and voltage to find the charge. The separation change does not affect the charge because the battery is disconnected.

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Key Concepts

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

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined as C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the plates. For a parallel-plate capacitor, capacitance can also be expressed as C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the plate area, and d is the separation between the plates.

Charge Conservation

Charge conservation is a fundamental principle stating that the total electric charge in an isolated system remains constant. When the battery is disconnected from the capacitor, the charge stored on the plates remains fixed, even if the separation between the plates changes. This means that the initial charge before the plates are pulled apart will equal the charge after the separation is increased.

Effect of Plate Separation on Capacitance

The capacitance of a parallel-plate capacitor is inversely proportional to the distance between the plates. As the separation increases, the capacitance decreases, which affects the voltage across the capacitor if the charge remains constant. When the plates are pulled apart, the voltage increases, and the relationship between charge, capacitance, and voltage must be considered to determine the new charge on the capacitor.

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Parallel Plate Capacitors

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Textbook Question

A parallel-plate capacitor with plate area 2.0 cm² and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor?

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