Prepared By Waqas Amin Sheikh He used a torsion balance For point charges ,charged bodies that are very small in comparison with the distance r between them He found that electric force is proportional to 1/r2 When the distance is doubles , the force decreases to ¼ of its initial value When the distance halved the force increases to four(4) times to its initial value The force between two point charges also depends upon the quantity of charge on each body To explore this dependence he…. Divided a charge into two equal parts by placing a small charged spherical conductor into contact with an identical but uncharged sphere By symmetry the charge is shared equally between the two spheres He found that forces that the two point charges q1 and q2 exert on each other are proportional to each charges and therefore are proportional to the product q1q2 of the two charges The magnitude of the electric force between two point charges is directly proportional to the product of the charges inversely proportional to the square of the distance between them If q1 and q2 are the two point charges separated by a distance r between them. Then according to coulomb’s law F α q1.q2 F α 1/r2 F = k|q1.q2|/r2 Where k is the constant of proportionality and its value is depend upon the system of unit used The absolute values bars are used in equation because the charges q1 and q2 can be either positive or negative While the force F will always remain positive When the both charges q1 and q2 have the same sign either positive or negative the forces are repulsive When the both charges q1 and q2 have the opposite sign positive and negative the forces are attractive The value of the constant k is depends upon the system of unit used In our study of electromagnetism we will use SI unit there fore the value of k K=8.987551787 * 109N.m2/C2 K=1/4πe0 1/4πe0=9.0*109 N.m2/C2 one coulomb =6*1018electrons Experiments shows that when two charges exerts a force simultaneously on a third charge, the total force acting on that charge is the vector sum of the forces that the two charges would exert individually When two electrically charged particles in empty space interact, how does each one know the other is there? What goes on in the space between them to communicate the effect of each one to the other? To introduce this concept look at the fig. Look at the repulsive forces of two positively charged bodies Let B has the charge q0 Let F0 be the electric force of A on B This is an force of “action at distance” that acts across empty space without needing any matter i.e( such as a push rod or rope) to transmit it through the intervening space We first envision that body A ,as a result of the charge that it carries ,somehow modifies the properties of the space around it. Then body B as a result of the charge that it carries, sense how space has been modified at its position The response of body B is to experience the force F0 To elaborate how this two stage process occur we first consider body A by itself we remove body B and label its former position as point P Charged body A produces or causes an electric field at point P. and all other points in the neighborhood. This electric field is present at P even if there is no other charge at P This is due to the charge of body A. If a point charge qo is then placed at point P it experience a force Fo We take a point of view that the force exerted on qo by the field at P Thus electric field is the intermediary through which A communicates its presence to qo The point charge qo would experience a force at any point in the neighborhood of A The electric field that A produces exists at all points in the region around A We can say that the point charge qo produces an electric field in the space around it and this electric field exerts the force – Fo on body A For each forces (force on A due to qo and on qo due to A) One charge sets up an electric field that exerts a force on the second charge This is an interaction between two charges A single charge created an electric field in the surroundings space , but this electric field cannot exert a The electric force on a charged body is exerted by the electric field created by the other charges bodies To find out there is an electric field we place a charged body which we call a test charge at the point(shown in fig. c) If the test charge experience an electric force then there is as electric field at the point Electric field E : the electric field E at a point as the electric force Fo experienced by a test charge qo . Thus the electric field at a certain point is equal to the electric force per unit charge experienced by a E = Fo/ qo The SI unit in which the unit of force is 1 N and the unit of charge is 1C the unit of electric field magnitude is 1 newton per coulomb If the field E is known at any point the electric force is given by Fo = E* qo Qo + Fo E Qo Fo - E The charge qo can be +ve or can be –ve . If qo is +ve the force will be in the same direction as E and vice versa caution : the electric force experienced by a test charge qo can vary from point to point so the electric field can also be different at different points Since the electric field E can vary from point to point it is not a single vector quantity but an infinite set of vectors quantities We call the location of the charge source point and the point P where we determine the field is called field point It is also useful to introduce a unit vector r that points along the line from source to the field point This unit vector is equal to the displacement vector r from source to the field point divided by the distance r between these two points i.e r =r/r if we place a small test charge qo at the field point P at a distance r from the source point the magnitude Fo of the force is given by the coulomb’s law Fo = 1/4πE |qqo|/r2 Magnitude of E of the electric field at P is E = 1/4πe |q|.r/r2 using the unit vector r . We can write a vector equation that gives both magnitude and direction of the electric field E = 1/4πe q.r/r2 Using equation below E = 1/4πe q.r/r2 we can measure the electric field caused by a single point charge But in most cases the electric field is distributed over space The charged plastic and glass rod have electric charge distributed over their surfaces Field caused by charge distribution we can imaElectric field calculation gine the distribution to be made up of many charges q1, q2, q3,……… At any point P each point charge produces its own electric field E 1, E 2, E 3,…. So a test charge qo placed at P experience a force F1= q0E 1 from charge q1 a force f2=q0E2 From charge q2 and so on The total force F0 = F1 + F2 +F3……..= q0E 1+ q0E 2…. combine effect of all the charges in the distribution is describe by the total electric field E at point P . From the def. of electric field E = Fo/ qo = E1 + E2 + E3…….. The total electric field at P is the vector sum of the field at P due to each point charge in the charge distribution this is the “ Principle of the superposition of electric field” The