Need a little help please Particle A of charge 2.00 times 10-4 C is at the origi

Question

Need a little help please

Particle A of charge 2.00 times 10-4 C is at the origin, particle B of charge -9.00 times 10-4 C is at (3.00 m, 0) and particle C of charge 2.00 times 10-4 C is at (0, 4.00 m). What is the x-component of the electric force exerted by A on C? What is the y-component of the force exerted by A on C? Find the magnitude of the force exerted by B on C. Calculate the x-component of the force exerted by B on C. Your response differs from the correct answer by more than 100%. N Calculate the y-component of the force exerted by B on C. Sum the two x-components to obtain the resultant x-component of the electric force acting on C. Your response differs from the correct answer by more than 10%. Double check your calculations. N Repeat part (f) for the y-component. Find the magnitude and direction of the resultant electric force acting on C. magnitude N direction degree counterclockwise from the +x-axis

Explanation / Answer

Distance between A and C is 3 m Distance between B and C is 5 m Formula for force = 9e9 q1q2/r^2 Use 9e9*ax1e-4*b*1e-4 = 90 ab =========================== a) Zero. b) y component of the electric force exerted by A on C 90*2*2 /4^2 = 22.5 N c) The magnitude of the force exerted by B on C. 90*-9*2/5^2 = 65.8 N d) Angle of BC with the x axis 180 – ? and tan ? = ¾ and hence 180 – ? = 143.13° the x component of the force exerted by B on C. is 65.8 cos 143.13 = -52.55 minus shows that it is toward left e). the y component of the force exerted by B on C. 65.8 sin 143.13 = 39.599 N f) Sum the two x components from part (a) and (d) = the resultant x component of the electric force acting along x -axis is -52.55. g) Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis is 22.5+39.59 = 62.09 N h) Resultant of -52.55 and 62.09 is v [(-52.55)*2 + (62.09)^2] = 81N Direction is at angle (90 + f) from the x axis.

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