Gauss’s Law

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Physics Electromagnetism Gauss’s Laws

	Application of Gauss’s Law to insulators and conductors:
	For insulators one of the following conditions must be satisfied: (i) The value of electric field can be argued by symmetry to be constant over the surface. (ii) The dot product in the surface integral of electric field can be expressed as a simple algebraic product EdA when E and dA are parallel to each other . (iii) The dot product of E and dA in the surface integral can be treated as zero when they are perpendicular to each other. (iv) The field can be argued to be zero over the surface. Consider some examples of the insulators: (i) For a point charge, the electric field is given by: (ii) For a spherically symmetric charge distribution: . Here, a is the radius of the charge distribution and r is radius of the Gaussian surface. (iii)For thin spherical shell, the electric field is: (for r > a) For conductors, the following points must be taken into account while calculating the electric field at a point. In addition we must consider the charge with no net motion inside the conductor so that the charge is in electrostatic equilibrium. A conductor in electrostatic equilibrium has the following properties: (i) The electric field is zero inside the conductor. (ii) If an isolated conductor carries a charge, the charge resides on its surface. (iii)The electric field just outside the charged conductor is perpendicular to the surface of the conductor and has a magnitude of σ/є₀, where σ is the surface charge density at that point. (iv)On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. Now let us consider some examples of the case of conductors: (i) For a conducting sphere of radius a carrying a net positive charge 2q inside a spherical shell the electric field is: E = 0 (for r < a) (for a < r < b) (for r > c) Here, a = radius of the sphere, b = inner radius of the shell and c = outer radius of the sphere.