LAS-GenPhysics2_MELC_14_Q3-Week-2

i General Physics 2 Activity Sheet Quarter 3 – MELC 14 Week 2 Electrostatics REGION VI – WESTERN VISAYAS SHS.

ii Development Team of Physical Science Activity Sheet Writer: Jessette L.Nolido Editor: Dannie Clark M. Uguil Schools Division Quality Assurance Team: Dannie Clark M. Uguil Rusil N. Sombito Eunice A. Malala Division of Negros Occidental Management Team: Marsette D. Sabbaluca Ma. Teresa P. Geroso Dennis G. Develos Zaldy H. Reliquias Raulito D. Dinaga Dannie Clark M. Uguil Othelo M. Beating Regional Management Team Ramir B. Uytico Pedro T. Escobarte, Jr. Elena P. Gonzaga Donald T. Genine Rovel R. Salcedo Moonyeen C. Rivera Anita S. Gubalane Minda L. Soldevilla Daisy L. Lopez Joseph M. Pagalaran General Physics 2 Activity Sheet No. 3 - Electrostatics First Edition, 2021 Published in the Philippines By the Department of Education Region 6 – Western Visayas Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. This Learning Activity Sheet is developed by DepEd Region 6 – Western Visayas. ALL RIGHTS RESERVED. No part of this learning resource may be reproduced or transmitted in any form or by any means electronic or mechanical without written permission from the DepEd Regional Office 6 – Western Visayas..

iii Introductory Message Welcome to General Physics 2! The Learning Activity Sheet is a product of the collaborative efforts of the Schools Division of Negros Occidental and DepEd Regional Office VI - Western Visayas through the Curriculum and Learning Management Division (CLMD). This is developed to guide the learning facilitators (teachers, parents and responsible adults) in helping the learners meet the standards set by the K to 12 Basic Education Curriculum. The Learning Activity Sheet is self-directed instructional materials aimed to guide the learners in accomplishing activities at their own pace and time using the contextualized resources in the community. This will also assist the learners in acquiring the lifelong learning skills, knowledge and attitudes for productivity and employment. For learning facilitator: The General Physics 2 Activity Sheet will help you facilitate the leaching- learning activities specified in each Most Essential Learning Competency (MELC) with minimal or no face-to-face encounter between you and learner. This will be made available to the learners with the references/links to ease the independent learning. For the learner: The General Physics 2 Activity Sheet is developed to help you continue learning even if you are not in school. This learning material provides you with meaningful and engaging activities for independent learning. Being an active learner, carefully read and understand the instructions then perform the activities and answer the assessments. This will be returned to your facilitator on the agreed schedule..

Name of Learner: ___________________________________________________ Grade and Section: ________________________________Date: ______________ GENERAL PHYSICS 2 ACTIVITY SHEET No. 3 Electrostatics I. Learning Competency with Code Solve problems involving electric charges, dipoles, forces, fields, and flux in contexts such as, but not limited to, systems of point charges, electrical breakdown of air, charged pendulums, electrostatic ink-jet printers. (STEM_GP12EMIIIb-14) II. Background Information for Learners Electricity has a definite niche in the modern world. Can you imagine your world without any gadget and appliances? What will be your reaction in the midst of your online class if suddenly you lose electricity during a brownout? You know the vital role played by electricity in today’s living. Human progress is dependent on electricity. In physics, electricity is a phenomenon associated with stationary or moving charge. When the charge is stationary, it creates electrostatic force on charged object; and when it is in motion, it creates magnetic effects. In this learning activity, you will be to apply concepts related to electrostatics in solving various quantities associated with electricity. III. Activity Proper Read and study carefully the sample problems. ACTIVITY 1: Electric charges o A quantity of electricity. It is a basic property of matter, occurs in discreet natural units o SI unit for charge is Coulomb which is mathematically represented as: q=I.t Where q= electric charge (C) I= electric current ( A) t= time (s) Source:https://www.google.com/search?q=charges&tbm=isch&ved.

5 Sample Problem: Current flow in 15 A exists in a copper wire for 15 seconds. How much the amount of charge that passes through the conductor at any area. Given: o Electric current (I) = 15 A o Time interval (t) =15 seconds Required? Charge (q) Solution: q=I.t q=(15A) (15s) q=225 C Electric field o Electric field is the region where electric force acts on charged body. o It is described as the electric force per unit charge. o SI unit of newton/coulomb (N/C). o A charged particle creates an electric field around it. If the particle is positively charged, the electric field lines point away from the charge; if the particle is negatively charged, the electric field lines point inward toward the charge. o An Electric field mathematically described as : Where: E is the electric field. F is a force. q is the charge. Where: E is the electric field. k=8.99x 109 Nm2/c2 q is the charge E = F /q E = 𝑘𝑟2 𝑞 Source:https://www.google.com/search?q=illustration+of+electric+field +with+label&rlz.

6 Sample Problem: A uniform electric field has a magnitude of 8 N/C is directed downward. Solve for the magnitude and direction of the force by a charge of -8C place in this field Given: E= 8 N/C directed downward q= -8C Solution: F=qE = (8C) (8N/C) =64 N o The charge is negative since the direction of the force is opposite in the electric field. Hence, the force of the charge must be directed upward. ELECTRIC FLUX o An Electric flux is a property of an electric field that described as the number of forces that intersect a given area. It is directly proportional to the number of electric field lines going through a virtual surface. o Electric flux can be expressed mathematically depending on the surface area: a. UNIFORM ELECTRIC FIELD - Represent the same magnitude of force and direction of charges regardless of its position in space Where: E = electric field (N/C) A = area of the surface Θ = is the angle between the electric field lines and the normal (perpendicular) Sample Problem: A concrete material has surface area of 0.500m2 exhibit an imaginary electric flux having a magnitude of 217Nm2/C in an angle of 300 to the area in contact. Determine the electric field in a given condition. Given: ΦE =217N.m2/C A= 0.500 m2 Θ =30 ΦE = E⋅A = EScosθ Source:https://www.google.com/search?q=electric+flux&tbm=isch&v ed=2ahUKEwidqLP_.

7 Solution: The electric flux which is passing through the surface is given by the equation as: E= Φ A(cosθ) E= (217N)m2/C 0.500 m2(cos30) E= 500 N.m b. NON-UNIFORM ELECTRIC FIELD- Represent different magnitude of forces on both side of the dipole Where: dΦ E = electric flux through a small surface area E⋅ = electric field dS= the component of area perpendicular to the field. Sample Problem: A non-uniform electric field given by E = 5.0x i + 3.0 j (in SI units) pierces the Gaussian cube as shown in the figure. What is the electric flux through the right face of the cube? Given: E = 5.0x i + 5.0 j From the given illustration x = 1 m towards the left direction x = 3 m towards right direction Solution: o Solve value of electric flux through the right face. Take note: Solve for the value of electric field on the right face. E = 5.0x i + 3.0 j o Substitute the value of x = 3m to the right E = 5.0(3) i + 3.0 j E = 15 i + 3 j o Solve for electric flux. You need to substitute the given information: φ = ∫ E·dA φ = ∫ (15 i + 3 j) · dA must be positive based on the direction of the flux piercing to the right direction Source:https://brainly.ph/question/274 2956 dΦE=E⋅dS.

8 o Then multiply dA to the terms inside the parenthesis, we have: φ = ∫ 15 dA o Multiplying with i produces 1 and with j produces zero. φ = 15 ∫dA The integral of dA is A φ = 15 A o In calculating area of the cube is A = (2)² = 4 m², substitute to the flux: φ = 15 (4m²) φ = 60 N·m²/C ELECTRIC FLUX (CLOSED SURFACE) Sample Problem: A piece of cylindrical conductor has a surface charge density of -15uC/m-2 with given radius of 0.30m. Solve for the following: a. Total charge that enclosed by the cylindrical conductor of radius 1.5m and length of 2m. b. Total electric flux that passes through the surface of the cylindrical conductor c. Electric field outside the cylindrical conductor 15m away from it. Given: Surface charge density = -15uC/m-2 Radius= 0.30m Solution: a. Solve for the total charge: Formula 𝑄 = 𝜎𝐴 Where: Q= charge A= area (lateral area) 𝜎= Surface charge density Substitute: 𝑄 = 𝜎𝐴 0r 𝜎(2𝜋𝑅𝑙) = (15x10-6 C/m2) (2𝜋) (0.3m)(2m) Q= 5.7 x10 -5 C b. Solve for the total flux : ϕ= 𝑄 ∈0 = 5.7x10-5C 8.85 x10-12 C2/N.m2 ϕ =6.4 x106 N.m2/C Source:https://www.google.com/search?q=formula+for+gauss+law.

9 c. Solve for electric field (outside the cylindrical conductor) E= 𝑅 𝜖0𝑟 E=15x10-6C (0.3m) 8.85 x10-12 C2/N.m2 (1.5m) E=3.38x105N/C ELECTRIC DIPOLE o Electric dipole consist of pair of charges of opposite sign (±q) separated by a distance d to be smaller compared to the distance from which the charges are normally detected in an electric field. o The product qd is a vector quantity which points from the −q charge to the +q charge. o The magnitude qd is known as the electric dipole moment for the pair, and is denoted p. Sample Problem: A compound particle carries a charge of +1x10-6m and -1x10-6m with a distance separation of 10cm.what is the electric dipole moment within this compound? Given: q= 1x10-6C d= 1x10-2m Solution: p=qd =(1x10-6C)(1x10-2m) p =1x10-8Cm Torque on a Dipole o It is a vector quantity which represent a measure of force that causes an object to rotate about an axis. o Mathematically expressed in: τ= pE sinΘ o Where p is the dipole moment, E is the electric field and θ is the angle between the moment arm and the electric field. Source:https://www.google.com/search?q=electric+dipole&tbm=isch &ved=2ahUKEwiTqNTQlbHuAhVN9pQKHXeAB_EQ2- https://www.google.com/search?q=electric+dipole+torque&tbm= isch&ved=.

10 Sample Problem Determine the maximum torque that the field can exert in a water molecules if it experiences a dipole moment of 6.1 x 10-3 C.m within a uniform field having the magnitude of 2 x 105 N/C. Given: p=6.1 x 10-3 C·m E= 2 x 10 5 N/C. Solution: τ= pE sinΘ =(6.1 x 10-3 C·m)( 2 x 105 N/C.) τ =12,200 N.m Guide Questions: o Problem 1. Solve for the maximum torque within a uniform electric field of 6 x 10 4N/C experiences a dipole moment of 5 x10 -14Cm. o Problem 2. A molecules of oxygen possess a charges of -2C and 2C separated from a distance of 0.2m.Calculate the dipole moment in this condition. o Problem 3 Solve for the electric flux that acts on a uniform electric field with a magnitude of 8000N/C approximately passes through a 10m2 flat surface area. o Problem 4. Determine the magnitude of an electric field at r=.0.01m from the center of non-conducting sphere with a radius R=0.001m possessing a charge of 7C which evenly distributed throughout the sphere.(rR). IV. Reflection Complete the statement below. I understand ______________________________________________________________ ______________________________________________________________ I don’t understand ______________________________________________________________ ______________________________________________________________ I need more information about ______________________________________________________________ ______________________________________________________________ Source:https://physics.gurumuda.net/electric-flux- problemas.htm.

11 V. Answer Key VI. References Gadong ,EarlySol and Belleza,Rafael, General Physics 2 (Philippines adaptation, Vibal Group 2016). Silverio ,Angelina A., Exploring life Through Science PHYSICS (Quezon City, Philippines) Phoenix Publishing House,2017 Reyes,Christopher G.Maed,General Physics 2for Senior High School. Great BooksTrading, 2018 Problem 1. 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝜏=pE Substitute 𝜏=(5 x10 -14Cm)( 6 x 10 4N/C) 𝜏 =3 X10-9Nm Problem 2. 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: p=qd Substitute p = qd. p = 2C x 0.02m p= 0.04 C-m. Problem 3. formula Φ =EAcos  Substitute: Φ = (C(m2) (cos 0) Φ=  x 104 Nm2/C Problem 4. formula: E = 𝑘𝑟2 𝑞 Substitute: E= 8.99x 109 Nm2/c2 ( 7C) (0.01m)2 E=6.3 x1014 N/C.