Clasificación de las plataformas digitales

15.1 2:3  10^ÿ8N.

15.2 Each of the charges ‡q exerts a force q²=…4p0a²† on the charge ÿq. The horizontal components of these forces are equal and opposite, and the net force on ÿq is down the page, as shown in Fig 15.44, with a magnitude

F ˆ 2q²

4p0a²cos 30^ˆ

3 p q² 4p0a²:

15.5 Outside a conducting sphere of radius R the potential varies as 1=r and may be written

…r† ˆ …R†  R=r. The electric ®eld E…r† is E…r† ˆ ÿ@

@r ˆR…R†

r² ,

which has its maximum value …R†=R at r ˆ R. If the corners of the box are rounded so that they are portions of spheres of radius R, the ®eld close to the corners is also …R†=R. The minimum radius Rmin that can be tolerated when the box is raised to a potential of 100 kV satis®es

15.9 There is a charge³₂Q on the outer surface of each plate. The charges on the inner surfaces are ‡¹₂Q for the

plate carrying a total charge 2Q and ÿ¹₂Q for the plate carrying a total charge ÿ¹₂Q.

15.10 The electric ®eld points outwards from the centre and at a distance x from the centre, within the slab, its magnitude is jxj=0.

15.11 The ¯ux of the electric ®eld at a distance r from the centre of the nucleus is due to the charge within r. If r is less than the radius R of the nucleus, and the charge density is , the total charge within a sphere of radius r is

43pr³. The area of this sphere is 4pr²and the ¯ux out of it is

4pr²E…r† ˆ4pr³

3₀ , giving E…r† ˆ r 3₀:

Outside the nucleus the electric ®eld is the same as for a point charge at the centre, that is, it is proportional to 1=r² and the maximum ®eld is at r ˆ R.

This local ®eld that exists close to the nucleus is enormous compared to the largest electric ®eld that can occur even over distances about the size of an atom. As indicated in Problem 15.5, the largest electric ®eld that is sustainable in air is only about 5  10⁶V m^ÿ1.

15.13 Assuming that the force between the balloons is the same as if each were a point charge, the charge on each balloon is 1:4 mC.

15.14 The ®eld on the axis is qa

 lies between the plates at a distance x from one of them. Calculate the induced charge on each plate.

15.31 The relative permittivity of water is 80.36 at 20^C and 60.76 at 80^C when measured in a steady electric

®eld. The density of water is 0:9982 g cm^ÿ3 at 20^C and 0:9718 g cm^ÿ3 at 80^C. Use these data to obtain the dipole moment and the polarizability of a water molecule.

The ®eld acts along the axis in the same direction, from the negative to the positive charge, on both sides of the dipole. For small a=r the magnitude of the ®eld is proportional to …1=r†³: it falls off faster with distance than the ®eld due to a point charge, because to second order in …a=r† the contributions of the positive and negative charges cancel out.

15.15 The three charges at BDE and the three at CFH are symmetrically placed with respect to the diagonal GA in Fig 15.45. By symmetry the force on the charge at A is therefore outwards in the direction GA from the opposite corner. The component of the force due to the charge at B along GA is

q² 4p0a² 1

p ,3

the component due to the charge at C is q²

and the force due to the charge at G is q²

4p0… 

p3 a†²:

The total force is therefore q² 15.16 For a charge at the centre of the cube the ¯ux of the electric ®eld out of one face is q=60. For a charge at one corner the ¯ux out of an opposite face is q=240.

15.18 The electrostatic energy is calculated by building up the charge on the nucleus from the centre, assembling thin spherical shells one by one like the successive layers of an onion. As in Problem 15.11, at a distance r from the centre of the nucleus, less than its radius R, the charge within r is⁴₃pr³. The potential at r is

1 4p0r4

3pr³:

The energy needed to bring from in®nity an extra thin shell of radius dr, carrying a charge 4pr²dr is

1 4p0r4

3pr³ 4pr²dr ˆ⁴³p²r⁴dr

0 ,

and the electrostatic potential energy of the whole nucleus is lead, Q ˆ 82e, giving a total electrostatic energy of 1:24  10^ÿ10J or, equivalently, 775 MeV.

15.19 The two dipoles attract one another, and their potential energy at a distance of 0.4 nm is ÿ0:126 eV.

15.21 The number of electrons is about 1300. Capaci-tors of about this size are used to store digits in dynamic random access memories (DRAMs).

15.24 When no dielectric is present, the electric ®eld E inside the capacitor is E ˆ V=d, where V is the voltage difference between the plates and d their separation. If their area is S, the charge Q on the plates is ₀ES and the capacitance C ˆ Q=V ˆ 0S=d.

When the space between the plates is half-®lled with an insulator, the ®eld inside the insulator is smaller than the ®eld outside by a factor equal to the relative permittivity . Call the electric ®eld at the surface of the plates E⁰. The ®eld inside the insulator is E⁰= and

The ratio of capacitances with and without insulator is

15.25 The electric ®eld at a point P on the axis is

E ˆ q

Making a binomial expansion in powers of a=r,

E ˆ q

to second order in …a=r†. For small …a=r† the ®eld from the charges falls off with distance as 1=r⁴, more rapidly than the ®eld of the dipole in Problem 15.14. The charges in this problem may be regarded as two dipoles close together, arranged so that their contributions almost cancel each other at large distances. Such an arrangement of charges is called an electric quadrupole.

15.27 The potential energy of one charge due to all the others is ÿe²ln 2=…2p0d†. For d ˆ 0:1 nm this is ÿ20:0 eV.

15.29 The potential due to the real charges ‡ per unit length and the imagined `image' charges ÿ per unit length is zero everywhere on the plane midway between them. The potential on the conducting plate is zero, and the induced charge must be distributed on this plate in such a way that the ®eld above the plate is the same as the ®eld due to the charges  per unit length. Inside the conductor the electric ®eld is actually zero, and the induced charges indeed move to the surface of the plane to ensure that this is so, as illustrated in Fig 15.46.

The electric ®eld at the surface of a conductor is always normal to the surface. To calculate the ®eld due to the line charge and the image charge we only need to consider the component normal to the

surface. At a distance y from the line joining , the normal component is The surface charge density  is related to the ®eld by E_?ˆ =₀(eqn (15.39)), and the surface charge density on the conductor at a distance y from the line joining the real and image charges is

 ˆ d

p…y²‡ d²†:

15.30 Let the potential at the plane occupied by the positive charge density be . This plane is at a distance x from one of the conductors; choose the origin of x to be at this plane. The electric ®eld between 0 and x is ÿ=x, and the induced charge density on the conductor at x ˆ 0 is ÿ0=x. Similarly, the induced charge density on the conductor at x ˆ d is ÿ0=…d ÿ x†. The total induced charge density must be ÿ since there is no electric ®eld outside the conductors. Hence

0 x ‡ 0

d ÿ xˆ ÿ and 0 ˆ x…x ÿ d†

2x ÿ d : Hence the charge density on the conductor at x ˆ 0 is ÿ…x ÿ d†=…2x ÿ d† and the charge density on the conductor at x ˆ d is ÿx=…2x ÿ d†.

Fig. 15.46

For a single charge or any number of charges at a distance x from one of the conductors, the division of the induced charge between the two conductors is the same as it is for a sheet of charge. X-rays and -rays cause the formation of electrons and ions when they interact with matter. Many X-ray and -ray detectors

work by collecting such electrons and ions by placing them in an electric ®eld. The electrons and ions move in opposite directions towards conducting plates. Induced charges appear on these conductors before the ions and electrons arrive, allowing the time of generating the ions and electrons to be determined accurately.