ELECTROSTATICS. kq q SUPERPOSITION

Transcription

1 ELECTROSTATICS As seen in class, we observe both attractive and repulsive forces. This requires at least two kinds of charge. Although more is possible we choose the simplest explanation consistent with experiment. We choose to call them + and in order to make Coulomb s Law give direction directly ( + *+ = +, -*- = +, +*- = -). The experimental result for the force between two charges is: F kq q 1 That the force should be proportional to q 1 q is required by Newton s second law. That it should be proportional to 1/R is required by Quantum Mechanics. Although we will not rigorously derive this we will give a physical motivation shortly. The constant k depends on the units chosen for charge. As noted in class the most natural choice would be the smallest charge nature provides. This leads to so called natural units in which the charge on an electron is -1 and on a proton is +1. There are many systems in use, but we will use the SI system exclusively. In this system we define k to be n- m /coul, where q is in coulombs. This results in the charge on a proton being coulombs. R R ˆ SUPERPOSITION The next question is whether electrostatics is linear, i.e., is the force of two charges on a third merely the sum of the 1 st on the third and the nd on the third. The experimental answer is yes provided the charges are not too large or too close together. More about this later. We will always assume these criterion are met and hence: where F 3 is the force on 3, F 13 F3 F13F3 the force on 3 due to 1, F 3 the force on 3 due to. ELECTRIC FIELDS Although it is in principle possible to do electrostatics with forces directly, it is much more convenient to introduce the concept of a field. For now this is purely a mathematical convenience. Later we will see that it has a fundamental importance.

2 The idea is to split the force on an object into two parts. The first is a disturbed environment. The second is the effect of that disturbance on the object. As noted in class it is like listening to a radio. The disturbance is always there (radiowaves), but you don t observe it until you turn on the radio. To experimentally determine the electric field, E, at a point we use the following definition: F q r E r lim q q0 where Fq r is the force on charge q located at r. This definition works fine for macroscopic objects, but since there is a smallest q (proton) we can t perform it on the atomic scale. We therefore recognize the reason for taking the limit and a achieve the desired result in a different way. We take the limit so that the disturbance of the point charge, q, will not disturb the field we are trying to measure. This could happen in two ways. The first is that q might feel the field it produces. The second is that it might cause the charges producing the field to move hence changing the field they produce at r. We overcome these problems by requiring that a point charge not produce a field on itself and by locking all other charges in place. Although this may be difficult in practice we can easily do it in theoretical calculations. GRAPHICAL REPRESENTATION OF FIELDS We now introduce an extremely useful method of visualizing fields. Although this takes a little effort to master, it will make things immensely easier, and is well worth the effort. Since force is a vector so is field. Thus far we have represented vectors by arrows where the length of the arrow gave the magnitude and the direction gave the direction of the vector. We now introduce a new method in which the direction is still given by the direction of a line at the point in question, but the magnitude is not related to the length of the line. Instead it is given by the number of lines/area perpendicular to the line. Field due to point charge

3 q F R kqqrˆ kq kq Er lim lim Rˆ Rˆ lim q0 q q0 qr q0r R Since the field is clearly isotropic (depends only on R) we represent it graphically by lines moving radially outward from the charge uniformly in all directions Now consider a sphere of radius R centered on Q. Since the surface of the sphere is perpendicular to the radius, and the lines are distributed uniformly in all directions, we have: N #/ area 4 R where N is the total number of lines. But the magnitude of E is equal to this: Since we already know E for a point charge: we must have: Thus: N E 4R kq E R kq N R 4 R

4 N 4 kq This is a fundamental result. Each point charge Q emits (+charge) or absorbs (-charge) 4πkQ lines. Later on when we look at Quantum Mechanics we will find a physical significance to these lines in terms of force particles. GAUSS S LAW We now make use of superposition to derive a fundamental result. Consider an arbitrary volume containing an arbitrary collection of charges. Since each charge acts independently of all the other (superposition), the total number of lines leaving the volume must be: #lines leaving 4 kq in In other words, it doesn t matter how the charges are arranged, the total number of lines leaving will be the same. The number/area at each point on the surface will depend on the arrangement, but not the total number leaving. INTEGRAL FORM OF GAUSS S LAW We need a method to calculate the number of lines leaving the volume. To do this we define a vector element of area as follows. Consider an infinitesimal element of area on a surface bounding a volume: We represent it by a vector of length da directed perpendicular to the surface and pointing outward.

5 We can now calculate the number of lines passing through this area. We know that E is the number of lines/area perpendicular to the lines. Hence we only get to count the component of da perpendicular to E. This is da E cos(θ). Hence the number of lines passing through da is # EdAcos But this is just: # EdA To find the total number of lines leaving the volume we must integrate this over the entire surface: We then have the integral form of Gauss s Law: #leaving E da 4koQin EdA It is often useful to define a permittivity of free space, ε 0, as: ko Then: Q EdA in o