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Chapter 25: Q30P (page 742)

How much energy is stored in 1.00m3of air due to the “fair weather” electric field of magnitude 150 V/m?

Short Answer

Expert verified

The energy stored in 1.00m3 of air due to the “fair weather” electric field of magnitude 150 V/m is 9.96×10-8J .

Step by step solution

01

The given data

a) Volume of air, V=1.00m3

b) Magnitude of electric field,E =150 V/m

02

Understanding the concept of the stored energy

We need to use the equation of the energy density and the energy stored per unit volume in the magnitude E of the electric field. Using the given formulae, we can get the required value of the stored energy.

Formulae:

The energy density of an electric field,u=12ε0E2 ...(i)

The energy stored per unit volume in the electric field,u=UV ...(ii)

03

Step 3: Calculation of the stored energy

Using equations (i) and (ii), we can get the energy stored in the electric field as follows: (ε0=8.85×10-12C2N.m2is permittivity of free space)

U=12ε0E2V=128.85×10-12C2N.m2150Vm21.00m3=9.96×10-8J

Hence, the value of the stored energy is 9.96×10-8J.

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Most popular questions from this chapter

In Fig. 25-28, find the equivalent capacitance of the combination. Assume that C1is10.0μF,C2is5.00μF,andC3is4.00μF

Figure shows a parallel plate capacitor with a plate areaA=5.56cm2and a separation ofd=5.56mm. The left half of the gap is filled with material of dielectric constantk1=7.00; the right half is filled with material of dielectric constantk2=12.0. What is the capacitance?

In figure, how much charge is stored on the parallel plate capacitors by the 12.0 Vbattery ? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10-3m2and a plate separation of 2.00 mm.

A slab of copper of thickness b=2.00mmis thrust into a parallelplatecapacitor of plate area A=2.40cm2and plate separationd=5.00mm , as shown in Fig. 25-57; the slab is exactly halfway between the plates.(a) What is the capacitance after the slab is introduced? (b) If a chargeq=3.40μCis maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in?

A parallel-plate capacitor has charge qand plate area A.(a) By finding the work needed to increase the plate separation from xto x = dx, determine the force between the plates. (Hint:See Eq. 8-22.) (b) Then show that the force per unit area (the electrostaticstress) acting on either plate is equal to the energy densityε0E2/2between the plates.
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