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Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory

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Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay We know that inside a conductor there is no electric field and that the surface of a metal is an equipotential surface. Further, the tangential component of the electric field on the surface is zero leaving with only a normal component of the field on the surface. Our argument in arriving at these properties was largely intuitive and was based on our expectation that under electrostatic conditions, we expect the system of charges to stabilize and move around. Let us look at this problem from a mathematical stand point. Consider surface of a conductor. Let us take the conductor surface to be in the xy plane. In the vacuum outside the conductor there are no free charges so that . Further as we are concerned with only electrostatic phenomena, everywhere. There are charges only on the surface of the conductor which would produce an electric field. The charges must be distributed in such a way that the fields they produce superpose so as to cancel the electric field inside the metal. So near the surface of the metal over a small a distance there is a change in the electric potential. Thus, if we take the outward normal to the surface of the conductor to be along the positive z direction, the z component of the electric field must be large near the surface. If the surface is homogeneous, or must be finite and continuous. Since the curl of the electric field is zero, this means, for instance, and since is finite and continuous, so must be , and by symmetry, . Since the field, and its tangential component, are zero inside the surface, the tangential components ( ) must be zero on the surface. Total charge on the surface of the conductor is obtained by integrating the surface charge density, which gives . Theorem : The potential function can be maximum only at the boundary of a region when the the electric field exists. Proof : Suppose, the theorem is not true and the potential is maximum at a point P which does not lie on the boundary of a region of electric field. If this were so, we could enclose the point P by a small volume at every point of which the potential is less than that at P. Thus everywhere on this region 1 . Thus the surface integral of this quantity . This contradicts the fact the according to Laplace’s equation this integral should be zero as the surface integral is equal to the amount of charge enclosed by the surface and away from the boundary there are no charges. Energy of a Charged Conductor in Electric Field We will calculate the energy that is stored in the field produced by a charged conductor. In doing so we will subtract the energy that is stored within the volume of the conductor itself. Let denote all space around the conductor which excludes the conductor volume itself. The stored energy is given by where we have used the identity The second integral on the right gives zero because outside the conductor there being no charge, The first integral can be converted to an integral over the surface using the divergence theorem. Since the volume excludes the conductor volume, the surface has two components, one at infinity, and the other at the surface of the conductor. The contribution to the surface integral from the surface at infinity is zero. We are then left with the integral over the surface of the conductor. Defining the normal direction on the conductor surface as being outward from the conductor surface, we observe that the boundary of the infinite region excluding the conductor volume has a normal direction which is opposite to the normal on the surface of the conductor. We thus have onductor the integral being over the surface of the conductor itself. If there are multiple conductors, we can extend the same argument and get the total energy stored in the field as 2 Where the potential has been taken out of the integral as the potential remains constant on each conductor surface. The surface integral above, by Gauss’s theorem is the amount of charge on the th conductor. We thus have, Since the field equations are linear, the potential and the charge are linear function of each other, the coefficient of proportionality can only depend on the shape of the conductor. Since, for a single conductor, we have, The coefficients are known as the coefficients of capacitance and are known as the coefficients of potential. In terms of these coefficients, the energy of a system of conductors can be written as It can be shown that these coefficients are symmetric under interchange of indices. To illustrate this, let us assume that we have a collection of conductors and we add a small charge only to the k-th conductor, keeping the charges on all other conductors the same. Let us look at the amount by which the total energy of the system changes. We have, 3 Where, in deriving the last step from its preceding step, we have changed the dummy summation index from I to j. However, we can get an equivalent expression for the change in the energy from the expression . It follows that if the charge on the k-th conductor is increased by , the energy would change by (Remember that the factor of ½ in the expression for energy came in order to avoid double counting of interactions between pairs. Since we are only interested in change in potential energy due to an additional charge being given to the k-th conductor, the change in the potential energy is simply given by the product of the additional charge with the potential at the site of the k-th conductor.) Comparing (1) and (2), we get We can also show that the diagonal term is the largest coefficient. Further, all coefficients are positive. Consider two conductors k and j. Suppose we bring in a unit positive charge from infinity to the k-th conductor, keeping all other conductors uncharged. By definition of the potential for a single conductor . Thus if is equal to the potential of the k-th conductor. Let us now bring the j-th conductor into the picture. Since k-th conductor has a net positive charge, field lines Arising out of this conductor will either terminate at infinity, or terminate on the j-th conductor from which the field lines will go to infinity. Since is the only charge, the potential on the j-th conductor is . The potential on the k-th conductor is still since . Thus, we have, , which gives, . 4 In Lecture 5 we proves Green’s theorem according to which, given to scalar fields and in a volume V bounded by a surface S, Let a potential field which arises from a volume charge distribution in V and a surface charge distribution on S. Likewise, let be a potential which arises from in V and on S. Using Maxwell’s equation, we have, and . Let the electric field corresponding to and be and respectively. We then have, Thus This is known as Green’s Reciprocity Theorem. Consider the volume to be bounded by conducting surfaces. Since the potential is constant on the conducting surfaces, we can write, where the sum is taken over the surface of the conductors. Example : A point charge q is located at a distance r from a spherical conductor of radius R (<r). What is the potential on the surface of the sphere? Since the only charge is the point charge at a distance r, we take volume charge density to be zero. We can consider the system as consisting of two conductors with the point charge being considered as a conductor with an infinitesimally small radius. If we put the charge on the surface of the sphere, the potential that it produces at the location of the charge at a distance r, is . If we use the reciprocity theorem, denoting the potential on the sphere when the charge is at the distance r from by , 5 which gives . (Note that this also follows from the definition of the coefficient of potential, Let the point charge be conductor 1 and the sphere as conductor 2. Let , I.e. let the sphere be given a charge q. The potential at conductor 1 due to the charge q being given to the sphere is . However, the potential at the conductor 1 due to the charge q on the sphere is . Thus . By symmetry of the coefficient of potential, . The potential on the sphere when the charge on the point conductor is then, We have here so that . Example 2 : Two conductors having capacitances and are placed apart at a distance r from each other. Calculate the capacitance of the pair. Let us put a charge q on conductor 1 keeping the conductor 2 uncharged. The potential of the conductor 1 is then and the potential on the conductor 2 due to the charge q at a distance r is . Since when and is when , from the relation , we have . Likewise, . Thus we have, If we look at it as a matrix equation, we have . We can invert this equation to express the charges in terms of the potential on the two conductors, 6 The inverse of the matrix is easily found to be Thus the coefficients of capacitances are given by, Capacitors are devices which are used to store charges. Consider a collection of a system of conductors of which a pair hold equal and opposite charge , the others being neutral.
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