STEM 306 Physics Midterms Reviewer PDF

ATOMIC Circle Xavier University Senior High School STEM 306 PHYSICS MIDTERMS REVIEWER [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 Electrostatics a branch of physics that deals with the phenomena and pro...

ATOMIC Circle Xavier University Senior High School STEM 306 PHYSICS MIDTERMS REVIEWER [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 Electrostatics a branch of physics that deals with the phenomena and properties of stationary or slow-moving electric charges. ELECTRICITY Flow of electrical power/charge Secondary energy source -- We obtain it through the conversion of other sources of energy also known as primary sources (coal, natural gases, etc.) Can be carried by wires and used for heat, light, and power for machines ELEMENTARY CHARGE Represented by “e” Charge carried by a single proton Fundamental physical constant Proton (+e) Electron (-e) [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTRICAL CHARGE Physical property of an object that causes it to be attracted toward or repelled by another charged object generates and is influenced by a force called an electromagnetic force. Can be positive (proton) or negative (electron) Represented by (q). SI Unit of charge: Coulomb (C) Named after the French Physicist: Charles - Augustin de Coulomb [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTRIC FIELD The space surrounding a charged body Any charged particle placed in it will experience an electrical force. Every charge are associated with one Michael Faraday English Chemist introduced the use of electrical lines of force to map out an electrical field. PROPERTIES OF ELECTRIC LINES OF FORCE Starts on positively charged particles and ends on negatively charged particles or infinitely. Either intersect or break as they pass from one charge to another The greater the number of lines of force, the stronger the electric field CONCEPTS The neutral point is the point where no lines of force pass The electric field is zero at the neutral point Neutral points are where the resultant field is subtractive, and the electric fields are equal but oppositely directly. The neutral between two like charges is a point between the two charges and nearer the smaller charge [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 CONCEPTS For two unlike charges, lines of force can pass from positive to negative charge The neutral point cannot be between them; it is an external point along the line joining them and nearer the smaller charge. Coulomb's Law Like charges repel and unlike charges attract Electrical force between two charged particles is directly proportional to the product of the magnitude of the charges between two charged particles. (Iq1q2I) And is indirectly proportional to the square of the distance between the charged particles. (r^2) Mathematically written as: WHERE: F = electrostatic force in newtons (N) k = Coulomb's constant (9*10^9 Nm ^2/C^2 ) q1 and q2= charges in coulombs, (C) r = distance between the charges in metres, (m) The absolute value of the product of q1 and q2, is determined because only the magnitude of the force is to be computed. The Force magnitude F is always positive. Electric force F is a vector quantity, which may be positive or negative depending on its direction. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SAMPLE PROBLEMS 1.A point charge (q1) has a magnitude of 3×10-6C. A second charge (q2) has a magnitude of -1.5×10-6C and is located 0.12m from the first charge. Determine the electrostatic force each charge exerts on the other. 2.Describe the changes to the magnitude and direction of the force on one of the charges in an electric dipole when the distance between the charges increases. -10 3.Two protons are separated by a distance 3.8×10 m in the air. Find the magnitude of the electric force if one proton exerts on the other, is the force attractive or repulsive? [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ANSWERS F = k q1q2 r 2 9 Nm 6 6 F = (9×10 C ) (3×10 C) (1.5×10 C) (0.12m) 2 -7 F = 2.81 × 10 N (Attractive) The magnitude of the force on each charge will decrease on the inverse square with distance, but the direction of the force will stay the same. F = k q1q 2 r 2 -19 -19 F = (9×10 C )(1.602×10 C)(1.602×10 C) 9 Nm (3.8×10 m) -10 2 F = 1.596×10 N (repulsive) -9 [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 Superposition Principle each charge will exert a force on another charge as if no other charged objects are present the net force that a particular charge experiences due to a collection of charges as a vector sum of all the individual charges. ELECTROSTATIC - TRANSFER Conductors - electric charge move freely Inductors - electric charge does not METHODS OF move freely Semiconductors CHARGING Superconductors CONTACT INDUCTION POLARISATION Two objects No contact Realignment of touching each required charge on other Grounding surface Conductors or source needed Contact or no insulators Conductors contact only Insulators only [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTRIC FIELD The Electric field (E) is a ratio of Force to charge It is a vector quantity with: - Magnitude: E=kqr2N/C - Direction: The direction of the force that charge (q) would exert on a small positive test charge placed in its vicinity WHERE: E = electric Field F = Electric force q = charge q0= test charge r = distance [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SAMPLE PROBLEMS 1.Two point charges, q1= 2.5 μC and q2= -4.0 μC, are placed 0.30 m apart. Determine the net electric field at a point P located 0.15 m from q1and 0.25 m from q 2. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ANSWERS Given: q 1 = 2.5 μC q = -4.0 μC 2 r1(distance from q1to P) = 0.15 m r2(distance from q2to P) = 0.25 m Required: Net electric field at point P. Equation: = kq r2 Net electric field: Enet= E1 + E2 Solution: Calculate the electric field due to q1: E1 = kq1/r12 = (9 × 109Nm2/C2)(2.5 × 10-6C) = 1 × 106N/C 2 (0.15 m) Calculate the electric field due to q2: E2 = kq2/r22 = (9 × 109Nm2/C2 )(-4.0 × 10-6C) = -5.76 × 105 N/C (0.25 m) 2 Calculate the net electric field: E net= E1 + E2 = 1 × 106 N/C - 5.76 × 105N/C = 4.24 × 105 N/C Answer: Therefore, the net electric field at point P is 4.24 × 10 5N/C. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTROSTATIC EQUILIBRIUM A state where an isolated conductor has no net motion of charge. Four properties: 1.Faraday demonstrated that the electric field is zero inside a closed conducting surface. 2.Any excess charge resides entirely on the surface of the conductor. 3.A room with a conducted frame protected Faraday from static charge, now known as Faraday cage. 4.The electric field just outside a charged conductor is perpendicular to the conductor’s surface. 5.On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature is smallest. ELECTRIC FLUX WHERE: “Flux” from the latin word “fluxus”, E is the magnitude of the means flow electric field Denoted as A is the area of the surface Total number of electric field lines the electric lines are passing passing a given area. through is the angle made by the area vector/direction of the plane and the axis parallel to the direction of flow of the electric field. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SAMPLE PROBLEMS 1.What is the electric flux through a sphere of radius 4m that contains a a.) +50uC at the centre? b.) -50uC charge at the centre? 2. Solve for the electric flux of the circular disk below: [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SOLUTIONS 1.) Given: ε₀ ≈ 8.85 × 10-12C²/Nm²) r = 4m a) +50uC at the centre b) -50uC charge at the centre Required: electric flux Equation: Φ = Qtotal / ε₀ Solution: a) +50 µC at the center: -6 Qtotal= +50 µC = 50 × 10 C ≈ 5.65 × 106Nm²/C (8.85 × 10-12 C²/Nm²) Φ = 5.65 × 106Nm² C b) -50 µC at the center: Qtotal= -50 µC = -50 × 10-6 C ≈ -5.65 × 106Nm²/C (8.85 × 10-12C²/Nm²) Φ = -5.65 × 10 6Nm² C [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ANSWERS 6 2 1. a.) =+5.65×10 Nm going outwards C b.) =-5.65×106 Nm2 going inwards. C 2. = 450 Nm2 C [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 GAUSS'S LAW Epsilon Naught Total electric flux through a surface is the total electric charge divided by Epsilon Naught Relates electric field, electric flux, and electric charge. Carl Fredrick Gauss; German scientist EQUATION Surface in Gauss’ law is called the Gaussian surface. Used to compute the electric field due to the system of point charges as well as continuous charge distribution. Charge distribution must be uniform and symmetrical. Expressed in terms of linear charge density (1d), Surface charge density (2d), and Volume Charge density. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SAMPLE PROBLEMS If a solid insulating sphere of radius 50.0 cm carries a total charge off 150 nC uniformly distributed throughout its volume, What is its (a.) Volume Charge Density, magnitude of its electric field at (b.) 10.0 cm and (c.) 65.0cm from its centre of its sphere. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 SOLUTIONS Given Radius (R) = 50.0 cm = 0.5 m Total charge (Q) = 150 nC = 150 × 10-9C Required: a. Volume charge density (ρ) b. Electric field (E) at 10.0 cm (r = 0.1 m) from the center c. Electric field (E) at 65.0 cm (r = 0.65 m) from the center Equation: a) ρ = Q = Q b) E = ρr c) E = kQ Solution: V ( 3 )πR³ 4 3×ε₀ r² a) ρ=Q= Q V ( 43 )πR³ ρ = (150 × 10-9C) ≈ 2.8 × 10-7 C/m³ ( 3 )πR³ 4 b) E = ρr 3×ε₀ E = (2.73 × 10-7 m³ C ) (0.1 m) ≈ 1080 N/C (3 × 8.85 × 10-12 Nm² C² ) c) E= Q ( 3 )πR³ 4 -9 E= 150 × 10 C ≈ 3190 N/C -12 C² (8.85 × 10 Nm² )4π (0.65 m)² [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ANSWERS A. 2.87×10-7 C/m 3 B. 1080 N/C C. 3191 N/C 3190 N/C [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTRIC CURRENT Movement of particles (Proton/electron); flow of electrons. Amount of charge passing through a conductor over time AMPERE WHERE: UNIT: Named after French scientist; Andre Mane Ampere I = Electric Columbs (C) / “Issac Newton of electricity” current second (s) or First to describe current as a Q = Charge Ampere (A); C/S continuous flow of electricity t = Time OR A along a wire Founded the science of CONVENTIONAL CURRENT electrodynamics. VS. ELECTRIC CURRENT 19th century: Benjamin Franklin introduced the terms: “Positive” and “Negative” charge. Conventional current Flow of positive charges from the positive terminal to the negative terminal of a source of voltage. Electric current Flow of electrons from the negative terminal to the positive terminal of a source of voltage. [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 OHM’S LAW States that the electric current is directly proportional to the Voltage across two points of a conductor, and is inversely proportional to the resistance. I=V/r where I = electric current, V = Voltage, r = resistance r =V/I where I = electric current, V = Voltage, r = resistance V=Ir where I = electric current, V = Voltage, r = resistance SAMPLE PROBLEMS 1. What is the voltage if a resistance of 25 Ω produces a current of 250 amperes? 2.What is the current produced by a voltage of 240 V through a resistance of 0.2 Ω? 3.What resistance would produce a current of 200 amps with a voltage of 2,000 V? ANSWERS [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 ELECTRIC POWER AND ENERGY Power expended in an appliance is the rate at which it consumes electrical energy. Obtained by multiplying voltage by its current; also through its resistance P=VI=I2 R=(V2 )/R Unit: Watt (W) When paying for the electric bills, you pay for the electric energy consumption (Eec), not the power. Energy consumption is determined by multiplying the power rating off the appliance by the length of time it was on. Unit: Joule (J); 1J = 1 watt-second Eec = P×t In bills, the energy consumption is shown as kilowatt-hour (kW-h) SAMPLE PROBLEMS 1.If the current and voltage of an electric circuit are given as 2.5A and 10V respectively. Calculate the electrical power. 2.Calculate the power of an electrical circuit consisting of resistance 3Ω and a current 4A flowing through this circuit? ANSWERS 1. 25 Watts 2. 48 Watts [email protected] Atomic Circle - AtCi GENERAL PHYSICS 2 Reviewer made by: Checked by: John Bryden L. Castillon BJ Budiongan STEM - 12 De Billy STEM - 12 Casati [email protected] Atomic Circle - AtCi