chapt04 - CHAPTER 4 ENERGY AND POTENTIAL In the previous...

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CHAPTER 4 ENERGY AND POTENTIAL In the previous two chapters we became acquainted with Coulomb's law and its use in finding the electric field about several simple distributions of charge, and also with Gauss's law and its application in determining the field about some symmetrical charge arrangements. The use of Gauss's law was invariably easier for these highly symmetrical distributions, because the problem of integration always disappeared when the proper closed surface was chosen. However, if we had attempted to find a slightly more complicated field, such as that of two unlike point charges separated by a small distance, we would have found it impossible to choose a suitable gaussian surface and obtain an answer. Coulomb's law, however, is more powerful and enables us to solve problems for which Gauss's law is not applicable. The application of Coulomb's law is laborious, detailed, and often quite complex, the reason for this being precisely the fact that the electric field intensity, a vector field, must be found directly from the charge distribution. Three different integrations are needed in general, one for each component, and the resolution of the vector into components usually adds to the complexity of the integrals. Certainly it would be desirable if we could find some as yet undefined scalar function with a single integration and then determine the electric field from this scalar by some simple straightforward procedure, such as differentiation. 83
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This scalar function does exist and is known as the potential or potential field . We shall find that it has a very real physical interpretation and is more familiar to most of us than is the electric field which it will be used to find. We should expect, then, to be equipped soon with a third method of finding electric fieldsÐa single scalar integration, although not always as simple as we might wish, followed by a pleasant differentiation. The remaining difficult portion of the task, the integration, we intend to remove in Chap. 7. 4.1 ENERGY EXPENDED IN MOVING A POINT CHARGE IN AN ELECTRIC FIELD The electric field intensity was defined as the force on a unit test charge at that point at which we wish to find the value of this vector field. If we attempt to move the test charge against the electric field, we have to exert a force equal and opposite to that exerted by the field, and this requires us to expend energy, or do work. If we wish to move the charge in the direction of the field, our energy expenditure turns out to be negative; we do not do the work, the field does. Suppose we wish to move a charge Q a distance d L in an electric field E . The force on Q due to the electric field is F E ± Q E ² 1 ³ where the subscript reminds us that this force is due to the field. The component of this force in the direction d L which we must overcome is F EL ± F Á a L ± Q E Á a L where a L ± a unit vector in the direction of d L : The force which we must apply is equal and opposite to the force due to the field, F appl ±´ Q E Á a L and our expenditure of energy is the product of the force and distance. That is,
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chapt04 - CHAPTER 4 ENERGY AND POTENTIAL In the previous...

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