Introduction to Electromagnetism - 6

Announcements There will be a quiz at the end of the lecture Review using some active figures Summary Main problem Given a set of electrical charges, how can we describe the electric phenomena caused by these charges We use the concepts of Electric Force Electric Fields and Field lines Electric Potentials (to come) The 4 rules for electric field lines 0 Example Question What does all this have to do with me and my program Answer The main objective is to understand biology better as a result of applying physical insight At its most fundamental level, biology is physics. Examples Electrostatic Equilibrium See Section 19.11 in textbook When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium Properties of a Conductor in Electrostatic Equilibrium Property 1 Einside 0 - Click to edit Master title style Click to edit Master text styles Second level Third level Fourth level Fifth level 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 6 January 26, 2007 Review q1 q2 Attractive Force if the charges have opposite sign ( and -) Repulsive Force if the charges have the same sign q1 q0 The force is between the 2 charges, but we can speak of one charge acting on the other q0 is always positive Therefore, observing the force acting on the test charge we can tell the sign and magnitude of q1 If the force acts so that the test charge goes away from q1, then q1 is a positive charge. If the force acts so that the test charge goes towards q1, then q1 is a negative charge. q1 q0 We have also seen that in reality the test charge is responding to a change in the electrostatic field caused by q1. (Remember the trampoline example) The Electric Field has the same direction of the Force that would act at that point on a positive test charge, and a magnitude equal to this force divided by the test charge. Moreover, if we know the Electric Field at any point, we can predict the force that would act on any charge q placed at that point. q1 If we have more that one charge generating the Electric Field, the net Electric Field at a point P can be found by summing vectorially the individual Electric Fields. q2 Of course, the Force that will be felt by a charge q placed at point P will be q1 If, instead of a discrete number of charges, we have a continuous distribution of charges (say linear) we treat the problem the same way using the concept of charge density. Continuous linear distribution of charges, TOTAL CHARGE Q LENGTH L Charge Density We know how to calculate the electric field at P due to charge q1 Let us use, instead of q1, a small charge dq contained within the differential length element dl We can then sum vectorially all the individual electric fields due to all the small elements that make up the line of charge Of course, since the charge distribution is continuous, the sum will become an integral. 6 39. A long, straight metal rod has a radius of 5.00 cm and a charge per unit length of 30.0 nC/m. Find the electric field (a) 3.00 cm, (b) 10.0 cm, and (c) 100 cm from the axis of the rod, where distances are measured perpendicular to the rod. 42. A solid copper sphere of radius 15.0 cm carries a charge of 40.0 nC. Find the electric field (a) 12.0 cm, (b) 17.0 cm, and (c) 75.0 cm from the center of the sphere. (d) What If How would your answers change if the sphere were hollow Table 24-1, p.754