Physics Lecture Notes 03, October 2024 PDF

October 13 – October 17 , 2024 (Third week) 03. Third lecture Electric force Electric field Electrical Energy and capacitance Electrical Energy and Cu...

October 13 – October 17 , 2024 (Third week) 03. Third lecture Electric force Electric field Electrical Energy and capacitance Electrical Energy and Current and Resistance Direct Current Circuits capacitance Magnetic field Induced voltages and inductance Dr. Mohamed Shaker Salem Alternating Current circuits Mechanical waves 1 Sound Waves  Gravitational field strength of earth (Remember) Let M is the mass that provides the gravitational field and m is thee mass that is affected by the field. a or g m rE ME At the surface of the earth N/kg = m/s² 2 A charged particle in an electric field When a particle of charge q and mass m is placed in an electric field E, the electric force exerted on the charge is qE. If this is the only force exerted on the particle, it must be the net force and so must cause the particle to accelerate. When Newton’s second law is applied to the particle, it gives - + + +q - - + - + - 3 + Accelerating a positive charge A body moving with a constant acceleration (a) a or g V= speed after a certain time t Vo= initial speed at t=0 X= position after a certain time t Xo= position at t=0 X- Xo= travelled distance 4  Problem A positive point charge q of mass m is released from rest in a uniform electric field E directed along the x axis, as shown in Figure. Describe its motion. Answer The charge moves with a constant acceleration due to the presence of a uniform, constant, electric field. Then, 5  (Remember) the free falling object in the earth’s gravitational field (Initial Position) (Final Position) Total Energy = constant KE=1/2 mv2= qEx KE+PE=constant 0+mgh=1/2 mv2+0 PE= qEx 1/2 mv2= mgh PE= qV 6 Change in electric potential energy  Coulomb force is a conservative force  The change in the electric potential energy, DPE, of a system consisting of an object of charge q moving through a displacement Dx in a constant electric field is given by 7 Electric potential difference  The electric potential difference, DV, between points A and B is the change in the electric potential energy as a charge q moves from A to B, divided by the charge q 8 Problem  Calculate the kinetic energy that an electron gains when accelerated through a potential difference of 1 V KE=1/2 mv2= qEx= qv KE= qV KE= eV KE= 1.6 x 10-19 J 9 The electron volt  The electron volt is defined as the kinetic energy that an electron gains when accelerated through a potential difference of 1 V 10 Electric potential created by a point charge q at a point in space a a r r - q +q q q va  k e v a  ke r r Electric potential created by multiple charges at a point Superposition principle The total electric potential at so me point a due to several point charges is the algebraic sum of the electric potentials due to the individual r1 a charges. q q q -q 1 r2 r3 v a  k e 1  ke 2  ke 3 r1 r2 r3 +q2 11 -q Electric potential of a pair of charges +q2 r q1 q1q 2 PE  k e r Potential energy of a system of charges +q2 r12 r23 +q1 +q3 r13 q1q2 q1q3 q2 q3 PE  ke  ke  ke r12 r13 r23 12  Problem A point charge q1 = 2 mC is located at the origin and a seconsd charge q2 = 6 mC is located at (0, 3.00) m. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4, 0) m. Find the potential energy of the two charges q1q2 PE  ke r12 6 6  6 X 10 X 2 X 10 PE  9 X 109 3 PE  0.036 J 13  Problem Find the total potential energy of the three charges shown in this figure q1q2 q1q3 q2 q3 PE  ke  ke  ke r12 r13 r23 PE  5.48 X 102 J 14  Problem Find the total electric potential due to these charges at the point P 15 The electric field due to a potential difference 16 Capacitance A capacitor consists of two parallel metal plates separated by a distance d. It is used in an electric circuit, the plates are connected to the positive and negative terminals of a battery. The capacitance, C, of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference between the conductors (plates): where eo is the permittivity of the free space 17 Computer Keyboard When a key on a computer keyboard is pressed, the spacing between the plates between the capacitor beneath the key decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. 18 Energy stored in a charged capacitor Problem A parallel-plate capacitor has an area A = 2 X 10-4 m2 and a plate separation d = 1 X 10-3 m. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a 3.00-V battery? (c) Calculate the charge density on the positive plate, assuming the density is uniform, and (d) the magnitude of the electric field between the plates. 19 Energy stored in a charged capacitor Problem A parallel-plate capacitor has an area A = 2 X 10-4 m2 and a plate separation d = 1 X 10-3 m. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a 3.00-V battery? (c) Calculate the charge density on the positive plate, assuming the density is uniform, and (d) the magnitude of the electric field between the plates. 20 Answer to the problem Or, 21 Capacitors in parallel Both capacitors have the same potential difference DV across them, because they are connected in parallel. 22 Capacitors in series For a series combination of capacitors, the magnitude of the charge must be the same on all the plates. 23 Problem (a) Calculate the equivalent capacitance of the capacitors shown in Figure. The capacitances are indicated in microfarads. (b) If a 12-V battery is connected across the system between points a and b, find the charge on the 4.0 mF capacitor in the first diagram and the voltage drop across it. 24 Answer 25 Capacitors with dielectrics A dielectric is an insulating material, such as rubber plastic, or waxed paper. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. If the dielectric completely fills the space between the plates, the capacitance is multiplied by the factor k, called the dielectric constant. (k > 1) e C k   1 eo C o 26 Capacitors with dielectrics For any given plate separation, there is a maximum electric field that can be produced in the dielectric before it breaks down and begins to conduct. This maximum electric field is called the dielectric strength 27 Problem A parallel-plate capacitor has plates 2.0 cm by 3.0 cm. The plates are separated by a 1.0-mm thickness of paper whose dielectric strength is 16X106 V/m. Find (a) the capacitance of this device, and (b) the maximum charge that can be placed on the capacitor. 28 Advantage of using dielectrics other than air in capacitors Insulating materials have values of k greater than unity and dielectric strengths greater than that of air. Therefore, a dielectric provides the following advantages: An increase in capacitance An increase in maximum operating voltage Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C 29