Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10

Question

Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10-4 C is at (7.00 m, 0) and particle C of charge 2.00 10-4 C is at (0, 4.00 m). (a) What is the x-component of the electric force exerted by A on C? N (b) What is the y-component of the force exerted by A on C? N (c) Find the magnitude of the force exerted by B on C. N (d) Calculate the x-component of the force exerted by B on C. N (e) Calculate the y-component of the force exerted by B on C. N (f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C. N (g) Repeat part (f) for the y-component. N (h) Find the magnitude and direction of the resultant electric force acting on C. magnitude N direction

Explanation / Answer

charge at point A is 6*10 -4 C charge at point B is -7 *10 -4 C charge at point C is 2*10 -4 C Force on charge at point C due to charge point A is FAC = k (6*10 -4 C ) ( 2*10 -4 C ) / 42 = ( 9*10 9 ) (6*10 -4 C ) ( 2*10 -4 C ) / 42 = 67.5 N x-component of the electric force exerted by A on C F ACx = FAC cos 90 = 0 y-component of the force exerted by A on C F ACy = FAC sin90 = F AC = 67.5 N Distance between B and C is x = ( 7 2 + 4 2 ) 1/2 = 8.06 m force exerted by B on C FBC = k (7*10 -4 C ) ( 2*10 -4 C ) / 42 = ( 9*10 9 ) (7*10 -4 C ) ( 2*10 -4 C ) / (8.06)2 = 19.3954 N angle ? = tan-1 ( 4/7 ) = 29.74 degrees x-component of the force exerted by B on C F BC x = F BC cos? = ( 19.3954 ) cos 29.74 = 16.8391 N y-component of the force exerted by B on C F BC y = F BC sin? = ( 19.3954 ) sin 29.74 = 9.6213 N x - component of force exerted on C By A abd B is F x = F ACx + F BC x = 16.8391 N y- component of force exerted on C By A abd B is Fy = F ACy - F BC y = 67.5 N - 9.6213 = 57.8786 N Resultant force F= ( Fx 2 + Fy2 ) 1/2 = 60.2784 N direction ? = tan -1 ( 57.8786 / 16.8391 ) = 73.77 degrees

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