Introduction to Electromagnetism - 4

Announcements Midterm I Friday, February 9 3 50-4 50 pm Room UA1120 Practice Problems for first 4 lectures have been posted Note blue problems have a higher degree of difficulty Review Electric Flux Electric Flux, General Area Electric Flux, Interpreting the Equation Electric Flux, General In the more general case, look at a small area element In general, this becomes Electric Flux, final The surface integral means the integral must be evaluated over the surface in question In general, the value of the flux will depend both on the field pattern and on the surface The units of electric flux will be N.m2/C2 Electric Flux, Closed Surface Flux Through Closed Surface, cont. Flux Through Closed Surface, final A positive point charge, q, is located at the center of a sphere of radius r The magnitude of the electric field everywhere on the surface of the sphere is E keq / r2 Conditions for a Gaussian Surface Field Due to a Point Charge Field Due to a Spherically Symmetric Charge Distribution Spherically Symmetric, cont. Spherically Symmetric Distribution, final Field Due to a Thin Spherical Shell Use spheres as the gaussian surfaces When r a, the charge inside the surface is Q and E keQ / r2 When r a, the charge inside the surface is 0 and E 0 Coulomb vs. Gauss A comparison Field at a Distance from a Line of Charge (Gauss) Field Due to a Line of Charge, cont. The end view confirms the field is perpendicular to the curved surface The field through the ends of the cylinder is 0 since the field is parallel to these surfaces Field Due to a Line of Charge, final Field Due to a Plane of Charge E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane Choose a small cylinder whose axis is perpendicular to the plane for the gaussian surface Field Due to a Plane of Charge, cont E is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinder The flux through each end of the cylinder is EA and so the total flux is 2EA Field Due to a Plane of Charge, final Click to edit Master title style Click to edit Master text styles Second level Third level Fourth level Fifth level Instructor Franco Gaspari PHY 1040U (Physics for the biosciences) Introduction to Electromagnetism and Optics Lecture 4 January 19, 2007 Continuous charge distribution Why do we need these definitions We know how to calculate the electric field due to a point charge. Using the concept of charge density, we can use the same method as for a point charge to calculate the electric fields due to a continuous distribution of charge. We then sum the contributions of all small elements to get the overall electric field Since the charge is distributed continuously, the sum is actually an integral. Area A Length L Total charge Q Total charge Q Total charge Q Volume V Electric Field Lines The electric field is a vector To visualize the electric field generated by a number of charges, we could draw a series of field vectors in several points in space This is not a very convenient method, since if we draw in few points, the representation is incomplete, if we draw in many points, we have a mess We need something simpler to visualize and describe Electric Fields. P1 P2 P3 P4 P5 we get a smooth curve This curve is called a field line. We will use field lines to give a representation of the electric field in space. D Electric field lines penetrating two surfaces. The magnitude of the field is greater on surface A than on surface B. Electric Field due to a line of charge Motion of a charged particle in a Uniform Electric Field Remember that we established that if we know the electric field, we know the force to which a charge q would be subject to if placed within that electric field, i.e., F qE Then The accelleration is in the direction of the Electric Field if q is positive. The accelleration is in the opposite direction if q is negative e- p Since the electric field is in the positive y direction, the electron is accelerated in the negative y direction. Before entering the electric field After entering the electric field At time t, then Electric Field Point charge Electric Field at point P Due to distribution of charges Force felt by charge q at P Due to an electric field Using vectors for the electric force is not useful (incomplete or messy). How do we visualize Field lines give better mapping (related to force via 4 rules). 8 dy Problem calculate By symmetry, we can consider of the rod and multiply by 2. Symmetry also tells us that we consider only the x component. dy Multiply x2 g Do these tomorrow ) L