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Chapter 19: Q48PE (page 698)

What charge is stored in the capacitor in Example 19.8?

Short Answer

Expert verified

The amount of charge stored in a capacitor when 3.00×103V is applied to it, is Q=26.6μC .

Step by step solution

01

Given data

Area of plates (A) = 1.00 m2

Separation between plates (d) = 1.00 mm

Voltage between the plates (V) = 3.00 x 103V

02

Concept Introduction

The ability of any capacitor to store energy in the form of electric field is known as Capacitance. The capacitanceC is given as –

C=QΔV.....................(1)

Here Q is the charge accumulated at the plates and ∆Vis the potential difference between the plates.

03

Calculation for Charge Stored

The charge stored in the capacitor is found by solving Equation (1) for Q –

Q=C×ΔV

Entering the values for C and ∆V, it is obtained –

Q=(8.85×10-9F)(3.00×103V)=26.6×10-6C=(26.6×10-6C)(106μC1C)=26.6μC

Therefore, the charge stored is Q=26.6μC.

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Most popular questions from this chapter

Construct Your Own Problem

Consider a heart defibrillator similar to that discussed in Example 19.11 . Construct a problem in which you examine the charge stored in the capacitor of a defibrillator as a function of stored energy. Among the things to be considered are the applied voltage and whether it should vary with energy to be delivered, the range of energies involved, and the capacitance of the defibrillator. You may also wish to consider the much smaller energy needed for defibrillation during open-heart surgery as a variation on this problem.

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(b) How many electrons per second flow if it takes \(5.00\)min to warm the formula? (Hint: Assume that the specific heat of baby formula is about the same as the specific heat of water.)

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