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M9L2

}Coulomb's Law in Two Dimensions: Equilateral Triangle To practice the vector addition of forces on charged objects in two dimensions, imagine three negatively charged points, each with a charge of -4.00 x 10 8 , fixed at the vertices of an equilateral triangle whose sides are 20.0 cm long. Find the net force on the top point. 1 2003 Since the charges are all of the same magnitude, the forces are all of the same magnitude. (4.00x 10C)(4.00x 10C) F =(8.99x10N- m"/C') - (0.200 m) F=3.596 Nor3.60 N To determine the direction of the net force, we will need to know the directions of the two forces acting on the top charge, and then add these two forces together by using the component method. Since the triangle is an equilateral triangle, the angles of the forces are at 60.0. The force on charge 1 by charge 3 is 3.60 N at an angle of 60.0 N of W. The x component of this force is -3.60 N cos 60.0 =-1.80 N. The y component of this force is 3.60 N sin 60.0 = 3.12 N. The force on charge 1 by charge 2 is 3.60 N at an angle of 60.0 N of E. The x component of this force is 3.60 N cos 60.0 = 1.80 N. The y component of this force is 3.60 N sin 60.0 = 3.12 N. The sum of the x components is -1.80 N + (1.80N) = 0.00 N. The sum of the y components is 3.12 N+ 3.12N = 6.24 N. Since there is no x component, the total force is 6.24 N north. In this example, we will examine the net force on one point charge due to three other charges. Four point charges A, B, C, and D, each with a magnitude of 4.0 x 10- C are found at the corners of a square whose sides are 20.0 cm long. Two of the charges are positive and two are negative as shown in the diagram below. Determine the net force acting on charge A. A (+) B (-) 20.0 cm D (+) The force on A due to B is FAB = F = (8.99x10' N. me/C2) (4.00 x 10-C)(4.00 x 10-C) = 3.60 N east. (0.200 m) The force on A due to C is FAC = F = (8.99x10'N. m /C2) (4.00 x 10 C)(4.00 x 10 6C) = 3.60 N south (0.200 m) To determine the force on A due to D, it is first necessary to know the distance between A and D. This distance is the hypotenuse of the triangle ABD. AD2 = AB2 + BD2 = (0.200 m) 2 + (0.200 m) 2 = 0.0800 m2 AD = 0.0800 m The force on A due to D is FAD = (8.99x10' N . m'/C2) (4.00 x 10-C)(4.00 x 10 6C) = 1.80 N Jo.0800 m? The direction of this force is 45.0 N of W.} Coulomb's Law in Two Dimensions: The Square: Adding the Forces To find the total x component of the net force, add together all of the x components. Fax = Fagx * Facx + Fapx = 3.60 N +0.00 N + (-1.80 N cos 45.0) =2.33 N To find the total y component of the net force, add together all of the y components. Fay = Fagy * Facy + Fapy =0.00 N + (-3.60 N) + 1.80 N sin 45.0 = -2.33 N The magnitude of the total force is F=JR+F 7 - f(2.33N)2 + (-2.33N) =330N The angle that this force vector makes is - 2.33N O=tan | == | =45.0S of E [2.331\\:] These examples illustrate that the method finding the net electrostatic force is to add together the force vectors produced by Coulomb's Law in the same manner that any vectors would be added together using the component method. Question(s): The physics of three charges in a plane. (Part 2) Three charges 91, 92, and 93 lie in a plane as shown below. 192 =-5.0 uC 123 = 0.20 m 45.0 91 = +4.0 uC.