FIELD OF A LINE CHARGE

FIELD OF A LINE CHARGE

Up to this point we have considered two types of 4 large distribution, the point charge and charge distributed throughout a volume with a density p” Clm3 If we now consider a filament like distribution of volume charge density, such as a very fine, sharp beam in a cathode-ray tube or a charged conductor of very small radius, we find it convenient to treat the charge as a line charge of density PL Clm. In the case of the electron beam the charges are in motion and it i~ true that we do not have an electrostatic problem. However, if the electron motion is steady and uniform (a de beam).

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and if we ignore for the moment the magnetic field which is produced, the electron beam may be considered to be composed of stationary electrons, for snapshots taken at any time will show the same charge distribution. Let us assume a straight-line charge extending along the z axis in a cylindrical coordinate system from -00 to bo,We desire the electric field intensity E at any and every point resulting from a uniform line charge density PI ” Symmetry should always be considered first in order to determine two specific factors: (I) )Vith which coordinates the field does not vary, and (2) which components of the field are not present. The answers to these questions then tell us which components.are present and wilh which coordinates they 40 vary. Referring  we realize that as we move around the line charge, varying 4> while keeping P and z constant, the line charge appears the same ..from every angle. In other words, aazimuthal symmetry is present, and no field component may vary, with  Again, if we maintain P and fjJ constant while moving up and down the line charge by changing z. the line charge still recedes into infinite distance in both directions and the problem is unchanged. This is axial symmetry and leads to fields which are riot functions of z.  If we maintain fjJ and z constant and vary p, the problem changes, and Coulomb’s law leads us to expect the field to become weaker as P increases. Hence, by a process of elimination we lfre led to the fact that the field varies only with p.

Now, which components are present? Each incremental length of line charge acts as a point charge and produces an incremental contribution to the electric field intensity which is directed away from the bit of charge ,(assuming a positive line charge). No element of charge produces a cJ> component of electric intensity; E”, is zero. However.each element does produce an Ep and n but the contribution to E; by elements of charge which are equal distances above and below the point at which we are determining the field will cancel. We.therefore have found that we have only an Ep component and it varies only with p. Now to find this component. Wea point P(O, y, 0) on the y axis at which to determine the field. This is a perfectly general point in view of the lack of variation of the field with cJ> and z. Applying (12) to find the incremental field atP due to the incremental charge dQ = ocdz’, we have

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This is the desired .answer, but there are many other ways of obtaining it. We might have used the angle 0 ‘as our variable of integration, for z’ = p cot fJ from and dz’ = – P csc20 dt). Since R = p csc f), our integral becomes, simply

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Here the integration was simpler, but some experience with problems of this type is necessary before we can unerringly choose the simplest variable of integration at the beginning of the problem.  We might also have considered (18) as our starting point.

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letting Pv dv’ = PL dz’ and integrating along the line which is now ‘our “volume” containing all the charge. Suppose we do-this and forget everything we have learned from the symmetry of the problem. Choose point P now at a general location (p, (), z).

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Before integrating a vector expression, we must know whether or not a vector under the integral sign (here the unit vectors a, and az) varies with the variable of integration (here dz’). If it does not, then it is a constant and may be removed from within the integral, leaving a scalar which may be integrated by normal methods. Our unit vectors, of course, cannot Change in magnitude, but a change in direction is just
as trouble some. Fortunately, t~ direction of ap does not change with t’ (nor with p, but it does change with  and az is constant always.

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Again we obtain the same answer, as we should, for there is nothing wrong with the method except that the integration was harder and there were two integrations to perform. This is the price we p:4- for neglecting the consideration of symmetry and plunging doggedly ahead with mathematics. Look before you integrate. Other methods for solving this basic problem will be discussed later after we introduce Gauss’s law and the concept of potential.  Now let us consider the answer itself.

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We note that the field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance. Moving ten times as far from a point charge, leads to a field only 1 percent the previous strength, but moving ten times as far from a .tine charge only reduces. the field to 10 percent of its former value. An analogy can be drawn with a source of illumination, for the light intensity from a point source of light also falls off inversely as the square of the distance to the source. The field of an infinitely long fluorescent tube thus decays inversely as the first power of the radial distance to the tube, and we should expect the light intensity about a finite-length tube to obey this law near the tube.

As our point recedes farther and farther from a finite-length tube, however, it eventually looks like a peint source, and the field obeys the inverse-square relationship. Before leaving this introductory look at the field of the infinite line charge, we should recognize the fact that not all line charges are located along the z axis. As an example, let us consider an infinite line charge parallel to the z axis at x = 6, y = 8, Figure 2.8. We wish to find E at the general field point P(x I y, z). We re lace p in (20) b the radial distance between the line charge and point.

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 We shall describe how fields may be sketched, and we will use the field of the line charge as one  Infinite uniform line’ charges of 5 nClm lie along the (positive and’ negative) x and y axes in free space. Find E at: (a) 1’A(O, 0, 4); (b) PB(O, 3, 4).

Ans. 458l VIm; 10.88y +36.98: V/m