Using the superposition principle the total electric eld E is the vector sum integral of these in nitesimal contributions Electric Field due to Continuous Charge Distributions E k q r r Pressures and Tensions Transmitted by Electric Fields Consider the general case where a closed surface an imaginary box is placed in an electric eld On the top face of the imaginary box there is an electric eld pointing in the outward direction of that face upward pull If we want to know the total electric force transmitted to the interior of this imaginary box in the up down direction we add these two transmitted stresses If there electric eld is homogeneous the force transmitted to the interior of the box in the up down direction is a pull upward plus an equal and opposite pull downward and adds to zero Textbook Readings Week 2 2 9 2 16 Volume Charge Density q V Surface Charge Density q A Line Charge Density q l Q r l Coulomb s Law Electric eld at a point in space Electric eld at a distance r from a charge q Using the superposition principle the electric eld due to a collection of point charges Acceleration of a particle of mass m and charge q moving in an electric eld The electric dipole moment vector p point from the negative to the positive charge and has a magnitude The torque acting on an electric dipole places in a uniform electric eld E is The potential energy of an electric dipole in a uniform external electric eld E is The electric eld at a point in space due to a continuous charge element q is Problem Solving Strategies Evaluating the vector integral E k q r r Start with E k q r r Rewrite the charge element q as Substitute q into the expression for E Specify an appropriate coordinate system and express the differential element and r in terms of the coordinates Rewrite E in terms of the integration variables and apply symmetry argument to identify non vanishing component s of the electric eld Complete the integration to obtain E 3 1 3 2 The number of electric eld lines that penetrates a given surface is called an electric ux EAcos The total ux through the entire surface can be obtained by summing over all the area elements E A Gauss s Law is expressed as E A Qenclosed where Qenclosed is the net charge inside the surface The application of Gauss s Law is limited only to systems that possess certain symmetry Cylindrical In nite Rod Coaxial Cylinder Planar In nite Plane Gaussian Pillbox Spherical Sphere or Spherical Shell Concentric Sphere Applying Gauss s Law Identify the symmetry associated with the charge distribution Determine the direction of the electric eld and a Gaussian surface on which the magnitude of the electric eld is constant over portions of the surface Divide the space into different regions associate with the charge distribution for each region calculate Qenclosed Calculate the electric ux through the Gaussian surface for each region Equation with Qenclosed and deduce the magnitude of the electric eld
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