Electrostatics PDF - PHY 102

ELECTROSTATICS PHY 102 General physics II 1  ELECTROSTATICS The interaction between static electric charges is called electrostatics. A stationary electrical charge that is built up on the surface of a material. The basis of electricity is...

ELECTROSTATICS PHY 102 General physics II 1  ELECTROSTATICS The interaction between static electric charges is called electrostatics. A stationary electrical charge that is built up on the surface of a material. The basis of electricity is charge. 2  Matter is made up of atoms. The charge on an atom is determined by the subatomic particles that make it up. Proton (positive charge) – + neutron (neutral) + + + electron (negative charge) – – – Atom Nucleus 3  Subatomic properties 1. Proton - has a positive charge and is located in the nucleus. 2. Neutron-has no charge (is neutral) and is also located in the nucleus as it fills in the spaces between the protons. 3. Electron- has a negative charge and is located outside of the nucleus in an electron cloud around the atom. 4  Points to note about charge The charged particles (electrons and protons) have the same magnitude of charge, which is e = 1.6 × 10-19 C where C is the Coulomb, the MKS unit of charge. The charge of a proton is +1.6 × 10-19 C and the charge of an electron is –1.6 × 10-19 C (the electron’s negative charge is just convention) 5  Lets do a little mathematics If you have 1 entire Coulomb of charge, how many electrons do you have? 6  Solution Because each electron has a charge of magnitude of 1.6 × 10-19 C, you have, Total number of electrons = 1 / 1.6 × 10-19 C = 6.25 × 1018 electrons 7  Subatomic Particle sizes A proton and neutron have about the same mass 1.67 × 10−27 𝑘𝑔 An electron has a much smaller mass 9.11 × 10−31 𝑘𝑔 8 How do atoms become “charged?” Atoms become charged when they become more positive or more negative. How can this happen? Remove or add a proton or an electron Protons and neutrons are bound together by the Strong Nuclear Force and it is very hard to separate them. Electrons, however, can be more easily removed 9 How do atoms become “charged?” When a balloon rubs a piece of wool... electrons are pulled from – + the wool to the balloon. + – The balloon has more – + – electrons than usual. – + The balloon: – charged, wool + The wool: +charged 10  Charge is a fundamental quantity like mass. Charge is denoted as q. Charge has a fundamental unit of a Coulomb (C). Charges are usually really really small numbers 11  Charge is a fundamental quantity like mass Electric charge always occurs as some integral multiple of a fundamental amount of charge ‘e’. In modern terms, the electric charge q is said to be quantized, where q is the standard symbol used for charge. That is, electric charge exists as discrete “packets,” and we can write q=Ne where “N” is some integer. 12 Multiples of Charges Chart 1e 1.6 x 10-19 2e 3.2 x 10-19 3e 4.8 x 10-19 4e 6.4 x 10-19 5e 8.0 x 10-19 13  Insulators and conductors Materials can be divided into two classes in electrostatics Conductors: materials that allow electrons to flow through them easily. (Metals, Human body) Conductors CANNOT be easily charged by friction as the extra electrons gained can easily escape. Insulators: materials that do NOT allow electrons to flow through them easily. Insulators can be easily charged by friction as the extra electrons gained CANNOT easily escape. 14  Insulators and conductors Electrons can be forced to move across an insulator by applying strong force (called electric field.) Then this acts like a conductor Semiconductors are a third class of materials, and their electrical properties are somewhere between those of insulators and those of conductors. 15  BECOMING CHARGED Materials can become charged in three ways Charging by friction Charging by induction Charging by conduction 16  Charging by induction: The production of a charge in an uncharged body by bringing a charged object close to it When negatively charged rod is put near a metal sphere... electrons of the sphere are pushed away from the rod. 17 18  Charging by conduction Consider a negatively charged rubber rod brought into contact with an insulated neutral conducting sphere. The excess electrons on the rod repel electrons on the sphere, creating local positive charges on the neutral sphere. On contact, some electrons on the rod are now able to move onto the sphere, neutralizing the positive charges. When the rod is removed, the sphere is left with a net negative charge. This process is referred to as charging by conduction 19 20  ELECTROSTATIC FORCE It is observed experimentally that a force exists between material objects that have the property of electric charge. It is the existence of this force that shows the existence of charge as one of the fundamental properties of matter along with size and mass. The electromagnetic force has two component parts: i. The electrostatic force which exists between stationary or “static” charges ii. The magnetic force which exists between moving charges or electric currents 21  Coulomb’s Law Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance, which he invented. From Coulomb’s experiments, we can generalize the following properties of the electric force between two stationary charged particles 22  The Electric Force is inversely proportional to the square of the separation r between the particles and directed along the line joining them; is proportional to the product of the charges 𝑞𝑎 and 𝑞𝑏 on the two particles; is attractive if the charges are of opposite sign and repulsive if the charges have the same sign; is a conservative force 23  ELECTROSTATIC FORCE This is a non-contact force (like the gravitational force except instead of two masses exerting force on each other the two objects charges exert a force of repulsion or attraction). ANY charged object can exert the electrostatic force upon other objects- both charged and uncharged objects. 24  ELECTROSTATIC FORCE 𝑞𝑎 𝑞𝑏 Where k is the proportionality constant 𝐹=𝑘 2 𝑞𝑎 𝑎𝑛𝑑 𝑞𝑏 are in coulombs (C) 𝑟𝑎𝑏 1 𝑘= ≅ 9.0 × 109 𝑁𝑚2 𝐶 −2 4𝜋𝜀0 𝜀0 is the permittivity of free space and has the value 𝜀0 = 8.85 × 10−12 𝐶 2 𝑁 −1 𝑚−2 25  Example 1 Consider three point charges located at the corners of a right triangle as shown below where 𝑞1 = 𝑞3 = 4.0𝜇𝐶, 𝑞2 = −1.5𝜇𝐶 , and a=0.10m. Find the magnitude of the resultant force exerted on q3 and the angle it makes with respect to the positive x-axis. 26 27  Solution The magnitude of the force exerted on 𝑞3 by 𝑞2 is 𝑞2 𝑞3 𝐹23 =𝑘 2 𝑎 −6 −6 9 2 −2 (1.5 × 10 𝐶)(4.0 × 10 𝐶) 𝐹23 = (9.0 × 10 𝑁𝑚 𝐶 ) (0.10𝑚)2 𝐹23 = 5.4𝑁 The magnitude of force exerted on 𝑞3 by 𝑞1 is (4.0 × 10 −6 𝐶)(4.0 × 10−6 𝐶) 𝐹13 = (9.0 × 109 𝑁𝑚2 𝐶 −2 ) 2(0.10𝑚)2 𝐹13 = 7.2𝑁 28  The resultant force exerted on 𝑞3 For the resultant force each of the Force will be resolved to x and y components before we can add the magnitudes. 𝐹3𝑥 = 𝐹13𝑥 + 𝐹23𝑥 = 5.1𝑁 + −5.4𝑁 = −0.3𝑁 𝐹3𝑦 = 𝐹13𝑦 + 𝐹23𝑦 = 5.1𝑁 + 0 = 5.1𝑁 2 2 𝐼𝐹 𝐼 = 𝐹3𝑥 + 𝐹3𝑦 = 5.12 𝑁 −1 5.12 𝜃 = tan = 86.6° 29 −0.3  Electric field ELECTRIC FIELD-is the environment created by an electric charge (source charge) in the space around it, such that if any other electric charges(test charges) is present in this space, it will come to know of its presence and exert a force on it. 30  INTENSITY (OR STRENGTH ) OF ELECTRIC FIELD AT A LOCATION The force on any particle with charge q placed at that point is 𝐹 = 𝑞𝐸 Electric Field (𝐸) is the force exerted on a unit charge when placed in an electric field. Given by 𝐹 𝑞 𝐸 = or = 𝑘 𝑞 𝑟2 where E is the electric field intensity. Its S.I. unit is V/M (volts per meter ) or N/C( Newton per Coulomb) 31 A test charge q0 at point P is at a distance r from a point charge q. (a) If q is positive, then the force on the test charge is directed away from q. (b) For the positive source charge, the electric field at P points radially outward from q. (c) If q is negative, then the force on the test charge is directed toward q. (d) For the negative source charge, the electric field at P points radially inward toward q. 32  Example A charge 𝑞1 = 7.0𝜇𝐶 is located at the origin, and a second charge 𝑞2 = − 5.0𝜇𝐶 is located on the x axis, 0.30m from the origin as shown above. Find the magnitude of the electric field at point P, which has coordinates (0, 0.40) m. 33 Solution 𝑞1 𝐸1 = 𝑘 2 𝑟1 7.0 × 10−6 𝐶 = 9.0 × 109 𝑁𝑚2 𝐶 −2 0.40𝑚 2 = 3.9 × 105 𝑁𝐶 −1 𝑞2 𝐸2 = 𝑘 𝑟22 5.0 × 10−6 𝐶 = 9.0 × 109 𝑁𝑚2 𝐶 −2 0.50𝑚 2 = 1.8 × 105 𝑁𝐶 −1 34  Solution The vector E1 has only a y component. The vector E2 has a x-component to be obtained by using 0.30 cos 𝜃 = = 0.60 0.50 Hence, 𝐸2𝑥 = 𝐸2 cos 𝜃 = 1.80 × 105 𝑁𝐶 −1 0.60 = 1.08 × 105 𝑁𝐶 −1 35 The vector E2 has a y-component also but a minus sign has to be Provided because the component is directed downwards to be obtained using 0.40 sin 𝜃 = = −0.80 0.50 Hence, 𝐸2𝑦 = 𝐸2 sin 𝜃 = 1.80 × 105 𝑁𝐶 −1 −0.80 = − 1.44 × 105 𝑁𝐶 −1 36  The resultant field at P is sum of 𝐸1 and 𝐸2 𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥 = 0 + 1.08 × 105 𝑁𝐶 −1 = 1.08 × 105 𝑁𝐶 −1 𝐸𝑦 = 𝐸1𝑦 + 𝐸2𝑦 = (3.90 × 105 − 1.44 × 105 )𝑁𝐶 −1 = 2.46 × 105 𝑁𝐶 −1 We use the Pythagorean theorem to find the magnitude of the resultant vector 𝐸= 𝐸𝑥2 + 𝐸𝑦2 = 2.72 × 105 𝑁𝐶 −1 The inverse tangent function yields the direction of the resultant vector 𝐸 5 𝑁𝐶 −1 𝑦 2.46 × 10 𝜙 = tan−1 = tan−1 = 66.6° 𝐸𝑥 1.08 × 105 𝑁𝐶 −1 37  Tutorial Questions 1. The magnitude of the electric force between two protons is 2.3× 10−26 𝑁. How far apart are they? 2. Estimate the magnitude of the electric field strength due to the proton in a hydrogen atom at a distance of 5.29× 10−11 𝑚 the expected position of the electron in the atom. 38 3. A charge of -4.00 nC is located at (0, 1.00) m. What is the x-component of the electric field at (4.00, -2.00) m? 39 1. The magnitude of the electric force between two 2 protons separated by distance r is 𝐹 = 𝑘 𝑒 𝑟 2 so the distance of separation must be 𝑘𝑒 2 (9.0×109 𝑁𝑚2 𝐶 −2 )(1.60×10−19 𝐶 r= = = 0.10𝑚 𝐹 2.3×10−26 𝑁 2. The magnitude of the electric field at distance r from a point charge q is E=𝑘𝑞 𝑟 2 so (9.0 × 109 𝑁𝑚2 𝐶 −2 )(1.60 × 10−19 𝐶) 𝐸= (5.29 × 10−11 𝑚)2 = 5.14 × 1011 𝑁𝐶 −1 40