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There is no industry today which does not need science to keep pumping red blood into its veins. Where research languishes, industry dies. Research is the life insurance of industrial progress. — David Sarnoff.

Consider again Example 5 of Chapter VIII. The potential at point A of Fig. 10-1 was found to be 20 ergs/esu. At point B the potential is

100 , 100 -186ergs


Suppose now that an electron is at rest at B. The force acting on it will be qE, where q is the charge of the electron and E the resultant electric field. This field can

, • i L j i- • 11 xu + 100 esu A +100 esn

be obtained by adding vectonally the Ch — io«n — >< — wcm — >o two fields Ei and E2.

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The force at position A is zero since the forces resulting from the two 100 esu charges are equal and in opposite directions.

If the electron is freed at B, it will experience a force in the direction B to A and so will travel along that path. When it gets to A the force acting on it will become zero. Let us determine what happens to the electron at A.

In dropping from B to A the electron loses potential energy, W.

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What happens to this potential energy? The first law of thermodynamics states that this energy must be conserved.

If the electron falls through a vacuum, there can be no frictional force and so no heat is developed. Let us say that the energy goes into kinetic energy (exceptions will be pointed out later).

The first law of thermodynamics then says:

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where m is the mass and v the velocity of the electron after falling through the potential difference (VA - Vb)- Let us see what this velocity is.

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(The mass of the electron is 9 x 10-28 gram. An experiment to measure this will be described in later work.) Solving for v,

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velocity of the electron as it passes point A will be 1.27 x 10* cm/sec in a direction away from B. The electron will then find itself moving against a retarding field and will have enough kinetic energy to reach a point 4 cm on the other side of A.

The important point to notice in the foregoing discussion is this: Electrical energy is no exception to the law of conservation of energy. If a body loses electrical potential energy, the energy must reappear in some other form; for a charge passing through a vacuum, it is most common for the energy to appear as kinetic energy.

Example 1. An electron emitted at zero velocity from a long straight filament of a radio vacuum tube is accelerated to a cylindrical plate B, 100 volts positive with respect to the filament (Fig. 10-2). What will be its velocity as it strikes the plate?

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100 Volts Vacuum tube.

Fig. 10

The loss of 100 volts potential will appear as kinetic energy.

300 esu 3 esu


Since the charge on the electron is 4.8 X 10~10 esu, the electron will lose

qV = 4.8 x 10-10 esu ^ 511 = 1.6 x 10~10 erg

By conservation of energy
qV = %rru?
1.6 X 10-10 = %x9X 10-28 t?

from which

v = 5.96 x 108 cm/sec

Notice here that the distance between the filament and plate did not enter the calculation. This is an important and very useful result of the potential method. (The acceleration of the electron is variable since the field is not constant in value.)

Example 2. An electron, moving with a velocity of v centimeters per second at an angle 0 with the horizontal, finds itself in an electric field at the point P of Fig. 10.3. The electric field is

due to a potential of V volts B c m < T

across the plates A B, separated '- v

a distance of d centimeters. a''

(a) What is the velocity of the electron as it strikes the plate?

(fo) Where does it strike the plate?

Solution. Let us first determine which plate the electron will strike. Since the lower plate is positive, the force on the electron is toward that plate; the electron, however, is given a vertical component of velocity of (v sin ff). This is similar to the gravitational problem of throwing a baseball in a room. If the vertical component of the velocity of the ball is great enough, it can overcome the pull of gravity. This will happen if

%m(v sin 0)2 ^ mgh [10-2]

where 0 is the angle of its initial velocity with the horizontal, m the mass of the ball, and h the height of the ceiling. Similarly for the electron, if


(where m is the electron's mass, e its charge, and V the potential difference between the point P and the upper plate), the electron will strike the upper plate.

If, however, %m(v sin 0)2 < eF'/300, the electron will not strike the upper plate but will proceed in a parabolic path (as in the baseball problem) toward the lower plate. The velocity of the electron as it strikes the lower plate can be easily computed from a consideration of the first law of thermodynamics.

According to the first law of thermodynamics, the total energy of the electron must remain constant. That is, the sum of the kinetic and potential energy of the electron at point P must equal the sum of the kinetic and potential energy at point n. Indicating the potentials at points P and n as VP and Vn respectively, and the velocities as vp and vn respectively,

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Since the electric field is uniform between the plates, this difference in potential (Vp - Vn) can be written

Then, if we know the mass of the electron m, its charge e, the initial velocity v, the distances s and d, and the potential difference V across the plates, the velocity «;„ as the electron strikes the lower plate can be computed.

The force on the electron, F, is constant, since the electric field is constant; it is given by



md 300

The task of finding the point where the electron strikes the plate is identical with the gravitational problem of the baseball. Placing the origin of the x and y coordinates at m, we have, for the y coordinate of the electron at any time t,

y = 2/0 + y sin 0 < + £a/ [10-6]

where ya = s and av - (eV/md300).

The time it takes the electron to strike the plate, t\, can be computed from equation 10-6 by setting y = 0. That is,


This being a second-degree equation, there will be two values for the time ti, one with a negative and the other with a positive value. The positive value is the time it takes the electron to go from P to n, and the negative value we interpret as the time it would have taken the electron to go from C to P along the same parabola.

We are now prepared to calculate where the electron will strike the lower plate, i.e, the distance from m to n, which we will call xn.

Since there is no acceleration in the x direction, we have

Xn = xo + (v COS 0)t

Here xa, the value of the abscissa at the starting point, is zero. At time t = ti, obtained from equation 10-7, the electron strikes the plate, and the value of xn is

Xn = (v COS 0)<i

which is the x coordinate of the point at which the electron strikes the plate; the y coordinate has (from our choice of axes) the value zero.

Relativity correction

Einstein has shown that the mass of a moving body increases appreciably as the velocity of the body approaches the velocity of light. Einstein's deduction has been verified experimentally; it is written

m = . TM° [10-8]

VI - (»'/<•2)

where m is the mass of a body moving at velocity v, wio is its rest mass, and c is the velocity of light. This correction in mass can usually be neglected for velocities less than 109 cm/sec; at this velocity the correction is less than 1 per cent. The kinetic energy of a moving body is expressed with the Einstein (or relativity) correction as


This correction for change in mass is usually called the "relativity correction" since it was first proposed by Einstein in his famous theory of relativity. Its use is imperative when dealing with high-speed electrons.

Measurement of electronic charge

A droplet of oil of mass m and charge -q is placed in an electric field, E, between two parallel plates of potential difference V (Fig.

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