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Chapter 16: Q19Q (page 663)

We discussed a method for measuring the dielectric constant by placing a slab of the material between the plates of a capacitor. Using this method, what would we get for the dielectric constant if we inserted a slab of metal (not quite touching the plates, of course)?

Short Answer

Expert verified

The value of dielectric constant of metals is infinite.

Step by step solution

01

Concept/Significance of dielectric.

The dielectric can be formed of a variety of materials, and it has an impact on the capacitor's quality, capacitance, and stability.

02

Determination of dielectric constant when a slab of metal inserted.

A metal is a conductor, it induces a charge on its surface that negates any field inside it. There will be no electric field between the capacitor plates if a metal slab is placed between conductor plates when a dielectric slab placed between plates the electric field will decrease that can be given by,

Edielectric=EK

Here, K is the dielectric constant.

The electric field of metal is given by,

Emeetal=EKEK=0K=E0=鈭�

Thus, the value of dielectric constant of metals is infinite.

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Most popular questions from this chapter

long thin metal wire with radius rand lengthLis surrounded by a concentric long narrow metal tube of radius R, whereR>>L, as shown in Figure 16.86. Insulating spokes hold the wire in the center of the tube and prevent electrical contact between the wire and the tube. A variable power supply is connected to the device as shown. There is a charge+Qon the inner wire and a charge鈭�Qon the outer tube. As we will see when we study Gauss鈥檚 law in a later chapter, the electric field inside the tube is contributed solely by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. (a) In terms of the chargeQ, lengthL, inner radiusr, and outer radiusR , what is the potential differenceVtube鈭�Vwire between the inner wire and the outer tube? Explain, and include checks on your answer. (b) The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. Calculate this power-supply voltage (give a numerical value), and explain your calculation. The lengthL=80鈥塩尘 , the inner radiusr=0.7鈥尘尘, and the outer radiusR=3鈥塩尘. This device is called a 鈥淕eiger鈥揗眉ller tube鈥� and was one of the first electronic particle detectors. The voltage is set just below the threshold for making the air glow near the wire. A charged particle that passes near the center wire can trigger breakdown in the air, leading to a large current that can be easily measured.

What is the potential (relative to infinity) at location B, a distance h from a ring of radius a with charge 鈥換 as shown in figure 16.94?

A small metal sphere of radius r initially has a charge q0 . Then a long copper wire is connected from this small sphere to a distant, large, uncharged metal sphere of radius R. Calculate the final charge q on the small sphere and the final charge on the large sphere. You may neglect the small amount of charge on the wire. What other approximations did you make? (Think about potential鈥�)

Location A is a distance d from a charged particle. Location B is a distance 2d from the particle. Which of the following statements are true? It may help to draw a diagram. (1) If the charge of the particle is negative, VB-VAis negative. (2) If the charge of the particle is positive, (VA<VB). (3) If VB<VA, we know that the particle must be positive. (4) VB<VA, regardless of the sign of the charge of the particle. (5) The sign of(VB-VA)does not give us any information about the sign of the charge of the particle.

LocationsA=<a,0,0>andB=<b,0,0>are on the +x axis, as shown in Figure 16.61. Four possible expressions for the electric field along the x axis are given below. For each expression for the electric field, select the correct expression (1鈥�8) for the potential differenceVA-VB. In each case K is a numerical constant with appropriate units.

(a)E鈫�=<Kx2,0,0>(b)E鈫�=<Kx3,0,0>(c)E鈫�=<Kx,0,0>(b)E鈫�=<Kx,0,0>(1)VA-VB=0(2)VA-VB=K(a-b)(3)VA-VB=K(1a-1b)(4)VA-VB=K(1a3a-1b3b)(5)VA-VB=12K(b2-a2)(6)VA-VB=KIn(ba)(7)VA-VB=K(a3-b3)(8)VA-VB=12K(1a2-1b2)
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