![]() Gauss's law immediately implies that the charge enclosed by the surface is equalto zero. Since the electric field inside the conductor isequal to zero, the electric flux through the Gaussian surface is equal to zero. Consider also aGaussian surface that completely surrounds the cavity (see for example thedashed line in Figure 2.14). For example,consider the conductor with a cavity shown in Figure 2.14. In this example we have looked at a symmetric system but the generalconclusions are also valid for an arbitrarily shaped conductor. The line integralof has to be evaluated for each of the three regions separately. Taking thereference point at infinity and setting the value of the electrostatic potentialto zero there we can calculate the electrostatic potential. The electrostatic potential V( r) canbe obtained by calculating the line integral of from infinity to a point a distance r from the origin. In the three different regions the electric fieldis equalto Since the conducting shell is neutral and any net charge must resideon the surface, the charge on the outside of the conducting shell must be equalto + q.ī) The electric field generated by this system can becalculated using Gauss's law. This can only be achieved ifthe charge accumulated on the inside of the conducting shell is equal to- q. However, according to Gauss's law this implies that thecharge enclosed by this surface is equal to zero. Therefore, the electric flux through any concentricspherical Gaussian surface of radius r ( a< r< b)is equal to zero. The potential difference between a and b is equal toĪ) The electric field inside the conducting shell is equal to zero(property 1 of conductors). The electrostatic potential V is constantthroughout the conductor.Ĭonsider two arbitrary points a and b inside a conductor (see Figure 2.12). Since the chargedensity inside a conductor is equal to zero, any net charge can only reside onthe surface.Ĥ. ![]() Anynet charge of a conductor resides on the surface. This immediately implies that the charge densityinside the conductor is equal to zero everywhere (Gauss's law).ģ. If the electric field inside a conductor is equal tozero, then the electric flux through any arbitrary closed surface inside theconductor is equal to zero. This property is a directresult of property 1. The chargedensity inside a conductor is equal to zero. Free charges will continue to flowuntil the cancellation of the initial field is complete.Ģ. If there would be an electric field inside theconductor, the free charges would move and produce an electric field of theirown opposite to the initial electric field. The electric field inside the conductoris equal to zero. Metallic conductors have the followingelectrostatic properties:ġ. In a metallic conductor one or more electrons per atom are free tomove around through the material. Check that your result is consistent with what youwould expect when z » d.ī) Repeat part a), only this timemake he right-hand charge -q instead of + q. Ī)ğind the electric field (magnitude and direction) adistance z above the midpoint between two equal charges q adistance d apart. Here is the unit vector from a segment of the charge distribution to the point at which we are evaluating the electric field, and r is the distancebetween this segment and point. To calculate the electric field at a point generated by these charge distributions we have to replace the summation overthe discrete charges with an integration over the continuous chargedistribution: volume charge ρ: the charge per unitvolume. surface charge σ: the chargeper unit area.ģ. line charge λ:the charge per unit length.Ģ. The following three differentdistributions will be used in this course:ġ. In most applications the source charges are not discrete, but aredistributed continuously over some region.
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