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1'... .9 V'-'J' gulps-surn- B. Coulomb's law provides an explicit expression for the electric field generated by a specific charge distribution: however. it has the

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1'... \".9 V""""'-'"J' gulps-surn- B. Coulomb's law provides an explicit expression for the electric field generated by a specific charge distribution: however. it has the limitation that: ta} It is a complicated method because it requires a totally vectorlal treatment. {bl It requires precise location of loads in space. to] It requires a referential surface that encloses the body under study. id) It requires knowledge of the magnitudes and locations of all leads contributing to E at a given point In space. T. Gauss's law states that: {a} The electric field flow must be symmetrical with respect to a surface through which the flow exits. 1b] The total electric field flux through a closed surface is equal to the net charge enclosed by the surface. to] The electric field flux generated by a volume ls equal on the surface enclosing lt. [d] The electric field flow ls conservative. B. A Gaussian surface should be chosen according to: {a} The coordinates to be used. lb) The geometry of the problem to be analyzed. to] The magnitude of the object to be analyzed. id} The amount of charge enclosed by the Gaussian surface. 9. What is the limitation of using Gauss's Law? {a} The constraint is that the object to be analyzed must be compatible with the selected Gaussian surface. lb) The limitation is that there must be symmetry in the object being analyzed. to} Sometimes it is difficult to choose the Gaussian surface. ldl There are not many Gaussian surfaces available. 10. If E ls zero at a given point In space. then this lmplles the absence of electric charges? ta} Not at all. since the absence of electric field may simply be the result of a field nullification by executing the superposition principle. {bl Of course. since an absence of loads inherently generates an absence of E as well. to] It Is not possible to be sure. until the calculations are done. id) If the potential is zero. then E Is also zero. regardless of whether there is an absence of charges

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