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Solution to Problem 5: The magnitude of an electric field due to a charge q is given by. E = k q / r2 and it is directed away from charge q if q is positive and towards charge q if q is negative. Hence the diagram below shows the direction of the fields due to all three charges.
If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.
If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.
Problem statement: Two point charges q 1 = q 2 = 10 -6 C are respectively located at the points of coordinates (-1, 0) y (1, 0) (the coordinates are expressed in meters). Calculate: The electric field due to the charges at a point P of coordinates (0, 1).
22 lis 2022 · Learn how to solve problems on electric field with clear explanations, examples, and exercises. This article is suitable for grade 12 and college students.
Practice Problems: Electric Fields. 1. (easy) What is the magnitude of a point charge whose E-field at a distance of 25 cm is 3.4 N/C? E= kq/r 2 3.4 = (9x10 9)q/(0.25) 2 q = 2.4x10-11 C. 2. (easy) A small charge (q = 6.0 mC) is found in a uniform E-field (E = 2.9 N/C). Determine the force on the charge. F = qE F = (6x10-3)(2.9) = 0.02 N. 3.
Examples of Electrostatic Problems with Dielectrics Problem: Find (electric flux density), (electric field intensity), and (polarization) for a metallic sphere (radius a, charge Q), coated by a dielec-tric (radius b), and the charge densities at the interfaces. Solution: Use Gauss’ Law In region 0, In region 1, a < r < b:
