Name: Grade and Section: Score: School: Teacher: Subject: General Physics 2 LAS Writer: JEREMIAH S. SABRIDO Content Evaluator: EMMA T. SURITA, CHRISTINE JOY G. SUA Lesson Topic: Electric Force and Electric Field Quarter 3 Wk. 1 LAS 2 Learning Target/s: 1. To calculate the net electric force on a point charge exerted by a system of point charges, and; 2. to determine the electric field due to a system of point charges using Coulomb's law and the superposition principle References: Serway and Faugn, J., 2006. Holt Physics. 10801 N. MoPac Expressway, Building 3, Austin, Texas 78759: Holt, Rinehart and Winston, pp.558-572. ELECTRIC FORCE AND ELECTRIC FIELD The electric force between two charges is given by Coulomb's Law. Coulomb's law states that the electric force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. It is given by Felectric = k-2 where, F = electric force; k = 8.98x109 Nm2/C2 (Coulomb constant); q1 = charge, q2= charge and; r= distance of separation. If there are multiple point charges, the forces add by superposition. According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges Ftotal= F12 + F1,3+F1,4. Fin). Consider a system of n charges, namely q1, q2, q3 .... qn. The force on q1 exerted by the charge q2, F12 = k 4192 Ap21 Here ^21 is the unit vector from q2 to q, along the line joining the two charges and r21 is the distance between the charges q, and q2- Sample Problem for electric force: Sample problem for Superposition Principle: The electron with a charge of 1.60x10-19 C and proton Consider three point charges located at the corners of a right triangle as shown in the figure below. Where q1 = q3 = 5.0x10-6 with 1.60x10-19 C of a hydrogen atom are separated by C, q2 = 2.0x10- C, and a = 0.10 m. Find the resultant force. a distance of approximately 5.3 x 1011 m. Find the magnitude of the electric to Solution: The force exerted by q1 on q3 is Given: F13. The force exerted by q2 on q3 qe= 1.60x10-19 C Felectric = k 9192 is F23. The resultant force F3 qp= 1.60x10-19 C Felectric = 8.99x109 (N.m?)/C exerted on q3 is the vector sum r= 5.3x10-11 m (1.60x10 19C)(1.60x10-19c) F13 + F23. k= (5.3x10-11m)2 Given: Solution: 8.99x109(N.m-)/C2 (8.99x109 (N.m2)/C2)(2.56x10-38C2) (5.3x10-11m) q2 = 2.0x10-6 F23 = k 9293 (2.30x10-28 N.m2) 93 = 5.0x10-6 = 8.99x109 (N.m?)/C2 x (2.0x10 C) (5.0x10 6c) (2.80x10-21m2) a = 0.10 m (0.10m)2 Felectric = 8.2x10-$ N k=8.99x109(N. (8.99x109 .m=) (1.0x10 11(2) 0.01m (8.99x10 20 (N.m) 0.01m2 F23 = 9.0 N ACTIVITY Solve the following problems. Use the rubrics as your guide in presenting your solution. Fully meeting expectations Minimally meeting Not meeting expectations expectations Solution Correct and complete solution with Incomplete solution with minimal Wrong solution no mathematical errors mathematical errors (1 point) (3 points (2 points) Final Answer Final answer is correct and labels Makes a small computation error | Answer is incorrect, or student it correctly or did not label it correctly does not have a final answer (2 points) (1 point) (0) 1-5. Two point charges, q1 = 8 UC and q2 = -5 JC, are separated by a distance r = 0.1 m. What is the magnitude of the electric force? Note: 1 JC = 10-6 C. 6-10. Three positive point charges of 3.0 uC, 6.0 UC, and 2.0 UC, respectively, are arranged in a triangular pattern, as shown in the figure below. Find the magnitude and direction of the electric force acting on the 3.0 nC 1.0 m 1.0 m-! 5.0 nC 1.0 m 2.0 nc