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Electrostatics May 25, 2017 iClicker An object is placed on the axis of a converging thin lens of focal length 2 cm, at a distance of 8 cm from the lens. The distance between the image and the lens is most nearly a. b. c. d. e. 0.4 cm 0.8 cm 1.6 cm 2.0 cm 2.7 cm Atomic Charges The atom has positive charge in the nucleus, located in the protons. The positive charge cannot move from the atom unless there is a nuclear reaction. The atom has negative charge in the electron cloud on the outside of the atom. Electrons can move from atom to atom without all that much difficulty. Charge Charge comes in two forms, which Ben Franklin designated as positive (+) and negative(–). Charge is quantized It only comes in packets The smallest possible stable charge, which we designate as e, is the magnitude of the charge on 1 electron or 1 proton. e is referred to as the “elementary” charge. e = 1.602 × 10-19 Coulombs. We say a proton has charge of e, and an electron has a charge of –e. Problem A certain static discharge delivers -0.5 Coulombs of electrical charge. How many electrons are in this discharge? Problem The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C. How many protons are in the system? How many electrons are in the system? Coulomb’s Law and Electrical Force Electric Force Charges exert forces on each other. Like charges (two positives, or two negatives) repel each other, resulting in a repulsive force. Opposite charges (a positive and a negative) attract each other, resulting in an attractive force. Coulomb’s Law Coulomb’s law tells us how the magnitude of the force between two particles varies with their charge and with the distance between them. k = 9.0 × 109 N m2 / C2 q1, q2 are charges (C) r is distance between the charges (m) F is force (N) Coulomb’s law applies directly only to spherically symmetric charges. Problem A point charge of positive 12.0 μC experiences an attractive force of 51 mN when it is placed 15 cm from another point charge. What is the other charge? iClicker Two isolated charges, + q and - 2q, are 2 centimeters apart. If F is the magnitude of the force acting on charge -2q, what are the magnitude and direction of the force acting on charge + q ? Magnitude (A) (1/2) F (B) 2 F (C) F (D) F (E) 2F Direction Toward charge - 2q Away from charge -2q Toward charge - 2q Away from charge - 2q Toward charge - 2q Superposition Electrical force, like all forces, is a vector quantity. If a charge is subjected to forces from more than one other charge, vector addition must be performed. Vector addition to find the resultant vector is sometimes called superposition. Problem Electric Fields The Electric Field The presence of + or – charge modifies empty space. This enables the electrical force to act on charged particles without actually touching them. We say that an “electric field” is created in the space around a charged particle or a configuration of charges. If a charged particle is placed in an electric field created by other charges, it will experience a force as a result of the field. We can easily calculate the electric force from the electric field. Electric Fields Forces exist only when two or more particles are present. Fields exist even if no force is present. Field Lines Around Charges The arrows in a field are not vectors, they are “lines of force”. The lines of force indicate the direction of the force on a positive charge placed in the field. Negative charges experience a force in the opposite direction. Field Line Rules Single Charges Double Charges Plate Charges Field Vectors from Field Lines The electric field at a given point is not the field line itself, but can be determined from the field line. The electric field vectors, and thus force vector, is always tangent to the line of force at that point. Force due to Electric Field The force on a charged particle placed in an electric field is easily calculated. F = Eq F: Force (N) E: Electric Field (N/C) q: Charge (C) Problem The electric field in a given region is 4000 N/C pointed toward the north. What is the force exerted on a 400 μg Styrofoam bead bearing 600 excess electrons when placed in the field? Ignore gravitational effects. iClicker An electron e and a proton p are simultaneously released from rest in a uniform electric field E. Assume that the particles are sufficiently far apart so that the only force acting on each particle after it is released is that due to the electric field. At a later time when the particles are still in the field, the electron and the proton will have the same a. b. c. d. e. direction of motion speed displacement magnitude of acceleration magnitude of force acting on them Problem A 400 mg Styrofoam bead has 600 excess electrons on its surface. What is the magnitude and direction of the electric field that will suspend the bead in midair? Problem A proton traveling at 440 m/s in the +x direction enters an electric field of magnitude 5400 N/C directed in the +y direction. Find the acceleration. Electric Fields The Electric Field surrounding a point charge or a spherical charge can be calculated by: E = kq / r2 E: Electric Field (N/C) k: 9 x 109 N m2/C2 q: Charge (C) r: distance from center of charge q (m) iClicker A point P is 0.50 meter from a point charge of 5.0 X 10-8 coulomb. The intensity of the electric field at point P is most nearly (A) 2.5 x 10-8 N/C (B) 2.5 x 101 N/C (C) 9.0 x 102 N/ C (D) 1.8 x 103 N/C (E) 7.5 x 108 N/C iClicker Charges + Q and - 4Q are situated as shown above. The net electric field is zero nearest which point? (A) A (B) B (C) C (D) D (E) E Superposition of Fields When more than one charge contributes to the electric field, the resultant electric field is the vector sum of the electric fields produced by the various charges. Again, as with force vectors, this is referred to as superposition. iClicker The diagram above shows an isolated, positive charge Q. Point (B) is twice as far away from Q as point A. The ratio of the electric field strength at point A to the electric field strength at point B is (A) 8 to 1 (B) 4 to 1 (C) 2 to 1 (D) 1 to 1 (E) 1 to 2 Electric Field Lines Electric field lines are NOT VECTORS, but may be used to derive the direction of electric field vectors at given points. The resulting vector gives the direction of the electric force on a positive charge placed in the field. Problem A particle bearing -5.0 μC is placed at -2.0 cm, and a particle bearing 5.0 μC is placed at 2.0 cm. What is the field at the origin? iClicker An electron is accelerated from rest for a time of 10-9 second by a uniform electric field that exerts a force of 8.0 x 10-15 Newton on the electron. What is the magnitude of the electric field? (A) 8.0 x 10-24 N/C (B) 9.1 x 10-22 N/C (C) 8.0 x 10-6 N/C (D) 2.0 x 10-5 N/C (E) 5.0 x 104 N/C iClicker An electron is accelerated from rest for a time of 10-9 second by a uniform electric field that exerts a force of 8.0 x 10-15 Newton on the electron. The speed of the electron after it has accelerated for the 10-9 second is most nearly (A) 101 m/s (B) 103 m/s (C) 105 m/s (D) 107 m/s (E) 109 m/s Problem A particle bearing 10.0 mC is placed at the origin, and a particle bearing 5.0 mC is placed at 1.0 m. Where is the field zero? Electric Potential Electric Potential Energy Electrical potential energy is the energy contained in a configuration of charges. Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. The unit is the Joule. Electric Potential Energy Electrical potential energy increases when charges are brought into less favorable configurations + - - Electric Potential Energy Electrical potential energy decreases when charges are brought into more favorable configurations. + + + - Work Work must be done on the charge to increase the electric potential energy. Electric Potential Electric potential is hard to understand, but easy to measure. We commonly call it “voltage”, and its unit is the Volt. 1 V = 1 J/C Electric potential is easily related to both the electric potential energy, and to the electric field. Electric Potential The change in potential energy is directly related to the change in voltage. ΔU = qΔV Δ U: change in electrical potential energy (J) q: charge moved (C) Δ V: potential difference (V) All charges will spontaneously go to lower potential energies if they are allowed to move. Electric Potential Since all charges try to decrease U, and ΔU = q ΔV, this means that spontaneous movement of charges result in negative ΔU. V=U/q Positive charges like to DECREASE their potential (ΔV < 0) Negative charges like to INCREASE their potential. (ΔV > 0) Problem A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change? E-field & Potential The electric potential is related in a simple way to a uniform electric field. ΔV = -Ed ΔV: change in electrical potential (V) E: Constant electric field strength (N/C or V/m) d: distance moved (m) Problem An electric field is parallel to the x-axis. What is its magnitude and direction of the electric field if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V? More Electric Field Stuff iClicker Two large, flat, parallel, conducting plates are 0.04 m apart, as shown above. The lower plate is at a potential of 2 V with respect to ground. The upper plate is at a potential of 10 V with respect to ground. Point P is located 0.01 m above the lower plate. The electric potential at point P is a. b. c. d. e. 10 V 8V 6V 4V 2V iClicker The electron volt is a measure of (A) Charge (B) Energy (C) Impulse (D) Momentum (E) velocity iClicker A point P is 0.50 meter from a point charge of 5.0 X 10-8 coulomb. The electric potential at point P is most nearly (A) 2.5 x l0-8 V (B) 2.5 x 101 V (C) 9.0 x 102 V (D) 1.8x 103 V (E) 7.5x 103 V iClicker The figure above shows two particles, each with a charge of +Q, that are located at the opposite corners of a square of side d. What is the direction of the net electric field at point P ? (A) Up and to the Right (B) Up and to the Left (C) Down and to the Right (D) Down and to the Left (E) Down Charge Location Electric field lines are more dense near a sharp point, indicating the electric field is more intense in such regions. All lightning rods take advantage of this by having a sharply pointed tip. During an electrical storm, the electric field at the tip becomes so intense that charge is given off into the atmosphere, discharging the area near a house at a steady rate and preventing a sudden blast of lightning. Excess Charges Excess charges reside on the surface of a charged conductor. If excess charges were found inside a conductor, they would repel one another until the charges were as far from each other as possible, thus the surface. E-Field of a Charged Sphere The electric field inside a conductor must be zero. + + + + + + E=0 + + + + + + + Conductor in an E-Field The electric field inside a conductor must be zero. - - + - + + E=0 - + + - + + iClicker A positive charge of 10-6 coulomb is placed on an insulated solid conducting sphere. Which of the following is true? (A) The charge resides uniformly throughout the sphere (B) The electric field inside the sphere is constant in magnitude, but not zero. (C) The electric field in the region surrounding the sphere increases with increasing distance from the sphere. (D) An insulated metal object acquires a net positive charge when brought near to, but not in contact with, the sphere. (E) When a second conducting sphere is connected by a conducting wire to the first sphere, charge is transferred until the electric potentials of the two spheres are equal. Electric Potential Energy iClicker Of the following phenomena, which provides the best evidence that particles can have wave properties? (A) The absorption of photons by electrons in an atom (B) The alpha-decay of radioactive nuclei (C) The interference pattern produced by neutrons incident on a crystal (D) The production of x-rays by electrons striking a metal target (E) The scattering of photons by electrons at rest Conservation of Energy In a conservative system, energy changes from one form of mechanical energy to another. When only the conservative electrostatic force is involved, a charged particle released from rest in an electric field will move so as to lose potential energy and gain an equivalent amount of kinetic energy. The change in electrical potential energy can be calculated by ΔU = qΔV. Problem If a proton is accelerated through a potential difference of -2,000 V, what is its change in potential energy? How fast will this proton be moving if it started at rest? iClicker Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is (A) 800 N/C (B) 1,600 N/C (C) 2,400 N/C (D) 5,000 N/C (E) 20,000 N/C Problem A proton at rest is released in a uniform electric field. How fast is it moving after it travels through a potential difference of -1200 V? How far has it moved? Electric Potential Energy Electric potential energy is a scalar, like all forms of energy. U = kq1q2/r U: electrical potential energy (J) k: 9 × 109 N m2 / C2 q1, q2 : charges (C) r: distance between centers (m) This formula only works for spherical charges or point charges. Electric Potential For a spherical or point charge, the electric potential can be calculated by the following formula V = kq/r V: potential (V) k: 9 × 109 N m2 / C2 q: charge (C) r: distance from the charge (m) Problem How far must the point charges q1 = +7.22 μC and q2 = -26.1 μC be separated for the electric potential energy of the system to be -126 J? Problem The electric potential 1.5 m from a point charge q is 2.8 x 104 V. What is the value of q? iClicker The hollow metal sphere shown above is positively charged. Point C is the center of the sphere and point P is any other point within the sphere. Which of the following is true of the electric field at these points? a. b. c. d. e. It is zero at both points. It is zero at C, but at P it is not zero and is directed inward. It is zero at C, but at P it is not zero and is directed outward. It is zero at P, but at C it is not zero. It is not zero at either point. E-Field and Potential Electric Field and Potential E=-V/d Two things about E and V The electric field points in the direction of decreasing electric potential. The electric field is always perpendicular to the equipotential surface. E-Field and Potential The electric field points in the direction of decreasing electric potential. E-Field is Perpendicular Zero work is done when a charge is moved perpendicular to an electric field W = Fd cos = 0 If zero work is done, and V = - W/q, then there is no change in potential. Thus, the potential is constant in a direction perpendicular to the electric field. Problem Draw a negative point charge of -Q and its associated electric field. Draw 4 equipotential surfaces such that ΔV is the same between the surfaces, and draw them at the correct relative locations. What do you observe about the spacing between the equipotential surfaces?