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Transcript
Coulomb’s Law 1 q1q2 FE 40 x 2 1 40 9 109 Newton meter 2 / coulomb 2 9 109 N m 2 / C 2 0 is the permittivity of free space Charge Charge q1 q2 Conservation of electric charge Charge is conserved: in any isolated system, the total charge cannot change. If it does change, then the system is not isolated: charge either went somewhere or came in from somewhere Principle of Superposition The presence of other charges does not change the force exerted by point charges. One can obtain the total force by adding or superimposing the forces exerted by each particle separately. Suppose we have a number N of charges scattered in some region. We want to calculate the force that all of these charges exert on some test charge q0 . q3 q2 q1 q7 q8 q5 q0 q4 q6 1 q0 q1rˆ1 q0 q2 rˆ2 1 2 2 ... Fq0 40 r1 r2 40 q0 qi rˆi . 2 ri i 1 N Problem 2 page 9 Two equal, positive charges of magnitude q are positioned along the x-axis as shown. What would be the force on a positive charge, q0 , placed on the y-axis a distance H from the x-axis? y q0 H q q a a x Principle of Superposition (revisited) The presence of other charges does not change the force exerted by point charges. One can obtain the total force by adding or superimposing the forces exerted by each particle separately. Suppose we have a number N of charges scattered in some region. We want to calculate the force that all of these charges exert on some test charge q0 . q3 q2 q1 q7 q8 q5 q0 q4 q6 1 q0 q1rˆ1 q0 q2 rˆ2 1 2 2 ... Fq0 40 r1 r2 40 q0 qi rˆi . 2 ri i 1 N 1 q0 q1rˆ1 q0 q2 rˆ2 1 2 2 ... Fq0 40 r1 r2 40 q0 qi rˆi . 2 ri i 1 N N We introduce the charge density or charge per unit volume q How do we calculate the total force acting on the test charge q? We chop the blob up into little chunks of volume V ; each chunk contains charge q V . Suppose there are N chunks, and we label each of them with some index . i q r̂i Let r̂i be the unit vector pointing from i th chunk to the test charge; let ri be the distance between chunk and test charge. The total force acting on the test charge is N 1 q( Vi )rˆi F 2 4 r i 1 0 i This is approximation! The approximation becomes exact if we let the number of chunks go to infinity and the volume of each chunk go to zero – the sum then becomes an integral: 1 qdV rˆ F 2 40 r V If the charge is smeared over a surface, then we integrate a surface charge density over the area of the surface A: 1 qdA rˆ F 2 4 r 0 A If the charge is smeared over a line, then we integrate a line charge density over the area of the length: 1 Qdl rˆ F 2 4 r 0 L Another example on force due to a uniform line charge A rod of length L has a total charge Q smeared uniformly over it. A test charge q is a distance a away from the rod’s midpoint. What is the force that the rod exerts on the test charge? dx ( x c) 2 3 2 x c( x c) 2 1 2 xdx ( x 2 c) 3 2 1 ( x 2 c) 1 2 The electric field y q3 q0 q4 x q1 q5 F E lim q0 0 q 0 E has the same direction as F q2 Newtons N E Coulomb C Michael Faraday 1791-1867 “The best experimentalist in the history of science” Electric field lines These are fictitious lines we sketch which point in the direction of the electric field. 1) The direction of E at any point is tangent to the line of force at that point. 2) The density of lines of force in any region is proportional to the magnitude of E in that region Lines never cross. How to calculate E ? 1) Put a “test charge” q0 at some point and do not allow it to move any other charges 2) Calculate the electric force on obtain E q0 and divide by q0 The force that N charges exert on a test charge 1 Fq0 40 q0: q0 qi rˆi . 2 ri i 1 N to We also calculated the force that a blob of charged material with charge density exerts on a test charge: 1 q0 dV rˆ F 2 40 r V We wrote the similar formulas if the charge is smeared out over a surface with surface density , or over a line with line density In all of these cases, the force ends up proportional to the test charge We might factor it out. This is the electric field! F E q0 . q0 . F E q0 qi rˆi 2 40 i 1 ri 1 V N 1 dV rˆ 40 r2 (N point charges) (Charge continuum) Given an electric field, we can calculate force exerted on some point charge q : F qE Example 1: Electric field of a point charge is directly radially away from or toward the charge. E 1 q ˆ r 2 40 r Example 2: Electric field of a dipole Have a great day! Hw: All Chapter 1 problems and exercises Reading: Chapter 1, 2