Christian Polytechnic Institute of Catanduanes, Inc. Francia Virac, Catanduanes GENERAL PHYSICS 2 SENIOR HIGH SCHOOL MODULE 1 ELECTRIC CHARGE, COULOMB’S LAW, ELECTRIC FIELDS AND ELECTRIC FLUX (Part 1) 7 1 7 1 Lesson 1: ELECTRIC CHARGE We can trace all electrical effects to electrons and protons inside every atom. This is because these particles have a property called electric charge. The electrons are negative and surround a dense, positive nucleus. Protons, which are positive, and neutrons reside in this nucleus. Neutrons are neutral and do not participate in electrostatic interactions. Just like mass, charge is fundamental property of subatomic particles. The smallest amount of charge is called the elementary charge, indicated universally by the symbol the element charge has magnitude of 1.60 x 10-19 C Particles proton Electron Neutron Mass (in kg) 9.1093897 x 10-31 kg 1.6726231 x 10-27 kg 1.6749286 x 10-27 kg Charge (in Coulomb (C)) +1.60217733 x 10-19 C -1.60217733 x 10-19 C None Location in atom Nucleus Outside nucleus nuclues The unit C stands for coulomb, named after French physicist Charles Agustin Coulomb. Example 1.1 How many electrons must an object lose so that it has a net positive charge of +1C? SOLUTION If the object loses one electron, its charge is +1.60 x 10 -19 C; if two electrons are lost, the object gains has a net charge of 2 x 1.60 x 10 -19 = 3.20 x 10 -19C And so forth. If we let n be the number of electrons lost and e the charge of each electron, then the total charge Q is Q = ne Thus, the number of electrons the object must lose so that it has a net positivity charge of +1C is or more than six quintillion electrons! This problem simply shows us that a charge of 1C is enormous. 7 1 7 1 Example 1.2 Aluminum has atomic number 13. This means that it has 13 electrons and 13 protons. What is the total charge of all electrons in an aluminum atom. SOLUTION We are given the number of electrons n. the charge of one electron is -1.60 x 10 -19 C. So if we have 13 electrons, each of which has a charge of -1.60 x 10 -19 C, then the total charge is: Q = ne = 13 (-1.60 x 10 -19 C) = -2.08 x 10 -18 C Thirteen electrons altogether still have very small charge Conductor and Insulator In certain materials such as aluminum, copper, and other metlas, the outermost or valence electrons are free to move around the entire materials, such materials are classified as conductor. In other materials, such as glass and wood, the electrons are more lightly bound to atom that they do not easily move around. These materials are insulators. good conductor are poor insulators poor conductors are good insulators. most metals are conductors while most nonmetals are insulators. In a conductor, the outermost electrons are farther away from the nucleus that they are more weakly bound to it. When atoms of a conductor form a bulk material, their outer electrons are no longer bound to the atoms but are free to float over the material. These are called free electrons. When energy is applied to these electrons, they can move in a more organized way, producing electric current. Free electrons can also be moved from one materials to another in process called charging. Removing an electron from an atom creates a positive ion. Charging In general, a material can be give a net charge by adding or removing electrons. There are several ways of doing this: 1. Using friction. By rubbing thighs together, electrons can peel off one material and remain in the other. 2. By touching. When a charged object comes into contact with another object, electrons are transferred, thereby charging the second object. 3. By induction. In this method, there is no actual contact between the charged object and that which is being charged. In this way, the charged object does not lose its charge to the object that gets charged in the process. 7 1 7 1 Using Friction By touching By induction An easy way to see charging by induction is by running a comb through your hair a few times (or rubbing it repeatedly with dry hands) and then holding it up over some small bits of paper. Lesson 2: COULOMB’S LAW In the previous lesson, we saw that like charges repel while unlike charges attract. This suggest the presence of the electrostatic force. This force is either a force of attraction between positive and a negative charges or a force of repulsion between two like charges. Electrostatic force holds the atom together. The positive nucleus attracts the negative electrons around it in a similar way that the Earth and other planets are held in orbit around the Sun by gravity. And this is not where the similarity ends. Recall that the force of gravity between two masses m1 and m2 is given by the equation: In which G is the gravitational constant and r is the distance between the centers of the masses. The electrostatic force between two charges Q1 and Q2 is given by a very similar equation. The magnitude of the electrostatic force Fe is: Where Fe is the electric force k is the Coulomb constant k = 8.9875 x 1097N · m2 / C12 or simplified by 9 x 109 N · m2 / C2 Q is the charge on object and r is the distance between the charges Take Note: If you encounter the nC (nanocoulomb) will show 10-9 when you convert the 109 coulomb. 7 1 This equation is called Coulomb’s law. It tells us that the force in newton’s between two charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. Note: Two things in Coulomb’s law. First, the absolute value sign means that the result is always positive. This is, after all, the magnitude of the force, so it has to be positive. Then, we see a constant k. This is the electrostatic constant. It is related to some fundamental constant of nature as follows: k= the constant ε0 is called the permittivity of free space and is equal to 8.85 x 10-12 coulomb squared per newton meter squared (C2/N · m2). The permittivity of a medium is a measure of its ability to store electrical charge. Substituting the constant needed to get the value of k, we get an approximately value of: k==k= 9 = 9 x 10 N · m2 / C2 Example 2.1 Two charge spheres Q1 = -2.00 x 10-10 C and Q2 = -5.0 x 10-8 C are held fixed at positions 5.00 cm apart. 1. Calculate the magnitude of each electrostatic force between the two spheres. Are the force attractive or repulsive? 2. If the first sphere is set loose, what will be its resulting acceleration? It has a mass of 5.00 x 10-3 kg. Solution 1. To solve for the magnitude of the electrostatic force Fe, we substitute the values into Equation 1: = 9 x 109 N · m2/ C2 = 3.6 x 10-5 N The electrostatic forces are repulsive because both charges are negative. 2. If the first sphere were to be released, then it will accelerate at a rate given by Newton’s second law of motion: Direction of Electrostatic Force The calculation we made ( Example 2.1) does not tell whether the force is directed toward the north, south, west, and east. The only way to find out is to sketch the problem out. 7 1 Let us assume that given positive charge Q1 and negative charge Q2, separately by distance r on a horizontal line. We already know that the force between them is attractive. In addition, the attraction is mutual. This means that Q1 pulls on Q2 7 1