Name: Grade & Section: Score: School: Teacher: Subject: General Physics 2 LAS Writer: NELSON B. MALIFICIADO Content Editor: RETCHIE JOY B PISANA, CHRISTINE JOY G. SUA Lesson Topic: Electric Potential and Electric Potential Energy Quarter 3 Week 2 LAS 3 Learning Target/s: 1. Relate the electric potential with work, potential energy, and electric field 2. Determine the electric potential function at any point due to highly symmetric continuous-charge distribution Reference(s): Villamor, R., 2003. Electronics Engineering. Sampaloc, Manila, Philippines: Ronnie Belarmino Villamor & HR Publishing, pp. 16. Electric Potential Difference and Electric Potential Energy The electrical potential energy U of a charge qiwith respect to another charge go at a distance r can be calculated using; 1 9091 U = ITTEO Where U has a unit of Joules (J); Joules, James Prescott (1818-1889), British physicist. Electric Potential Difference (V) is the amount of work needed to take a 1 coulomb charge from one point to another, or the potential energy per unit charge. Its unit is Volt, named after the Italian physicist, Alessandro Volta (1745-1827). U (Joules) V = - q (Coulombs) The electrostatic potential, V due to a 1 charge q at any distance r can be found by: V = - V = k- ATEOr Volts In terms of electric field E: Potential difference between two points in a uniform E field. V = Er V (volts) VAB = VA - VB = K_ -ka E =- r (meter) Where ra and r, are the distances of points A and B from Q Sample Problem A charge moves a distance of 2.0 cm in the direction of a uniform electric field whose magnitude is 215 N/C. As the charge moves, its electrical potential energy decreases by 6.9 _10-19 J. Find the charge on the moving particle. What is the potential difference between the two locations? Given: AU = -6.9x10-19] r = 0.020 m E = 215 N/C Unknown: q =? AV =? Solution: Use the equation for the change in electrical potential energy. AU = -qEr Rearrange to solve for the charge in electrical potential energy. AU (-6.9x10-19]) q = - Er q = 1.6 x 10-19 C -CN) (0.020 m) The potential difference is the magnitude of E times the displacement. AV = -Er = - (215 -) (0.020 m) AV = -4.3V Activity Solve the following problems. 1. As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 x 10-19J. What is the particle's charge? 2. What is the potential difference between the initial and final locations of the particle in Problem 1? 3. An electron moves 4.5 m in the direction of an electric field of strength 325 N/C. Determine the change in electrical potential energy