This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 23.47 m A point charge q1 = 4.00 nC is placed at the origin,
and a second point charge q; = 3.00 ml: is placed on the .raxis
at x = +201) cm. A third point charge q3 = 2.001s: is to be
placed on the x—axis between ql and (11. (Take as zero the potential
energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges
if qg is placed at x = +101) cm?I (b) Where should qg, be placed to
make the potential energy of the system equal to zero? 23.51 .. Determining the Size of the Nucleus. When
radium226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon222. We can
model this decay by thinking of the radium226 as consisting of an alpha particle emitted from the surface of the spherically sym—
metric radon222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.?9 Mel»r when the
alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The
radon nucleus contains 36 protons, while the alpha particle has 2 protons and the radium nucleus has 33 protons. (a) What was the electric potential energy of the alpha—radon combination just
before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus. 23.55  CALC A vacuum tube diode consists of concentric cy
lindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the elec
tric potential between the electrodes is given by V(J:) = Cx4’r3 where x is the distance from the cathode and C is a constant,
characteristic of a particular diode and operating conditions.
Assume that the distance between the cathode and anode is
13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of C. (b) Obtain a formula for the electric
field between the electrodes as a function of x. (c) Determine the force on an electron when the electron is halfwayr between the
electrodes. 23.65 " Electrostatic precipitotors
use electric forces to remove pollutant
particles from smoke, in particular in the smokestacks of coalburning power plants. One form of precipitator consists
of a vertical, hollow, metal cylinder with
a thin wire, insulated from the cylinder,
running along its axis (Fig. P2165). A
large potential difference is established
between the wire and the outer cylin
der, with the wire at lower potential. This sets up a strong radial electric
ﬁeld directed inward. The field pro duces a region of ionized air near the wire. Smoke enters the precipitator at
the bottom, ash and dust in it pick up 50.0 kV Figure P2165 electrons, and the charged pollutants are accelerated toward the
outer cylinder wall by the electric field. Suppose the radius of the central wire is 90.0 nm, the radius of the cylinder is 14.0 cm, and a potential difference of 50.0 kV is established between the wire
and the cylinder. Also assume that the wire and cylinder are both very long in comparison to the cylinder radius, so the results of
Problem 23.61 apply. (a) What is the magnitude of the electric ﬁeld
midway between the wire and the cylinder wall? {b} What magni
tude of charge must a 30.0ng ash particle have if the electric field
computed in part (a) is to exert a force ten times the weight of the particle? 23.82  CALC A hollow, thinwalled insulating cylinder of ra
dius R and length L (like the cardboard tube in a roll of toilet paper)
has charge Q uniformly distributed over its surface. (a) Calculate
the electric potential at all points along the axis of the tube. Take
the origin to be at the center of the tube, and take the potential to
be zero at infinity. (b) Show that if L is: R, the result of part (a)
reduces to the potential on the axis of a ring of charge of radius R. (See Example 23.11 in Section 23.3.) (c) Use the result of part (a)
to find the electric field at all points along the axis of the tube. ...
View
Full Document
 Spring '18
 Electron, Charge, Electric Potential, Alpha particle, Qg, 3.00 mL, 13.0 mm

Click to edit the document details