Particle A of charge 3.03 x 10^4 C is at the origin, particle B of charge -6.30
ID: 1594634 • Letter: P
Question
Particle A of charge 3.03 x 10^4 C is at the origin, particle B of charge -6.30 x 10^4 C is at (4.00 m, 0), and particle C of charge 1.07 x 10^4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.Explanation / Answer
(a) A and C are both positive ... so they repel each other
The force exerted by A on C [F(AC)] is along the y-axis so has no x-component
so x-component of F(AC) = 0 N
(b) y-component of F(AC) = [9 x 10^9 x 3.03 x 10^(-4) x 1.07 x 10^(-4)] / 3.00² = 32.421 N
(c) magnitude of F(BC) = [9 x 10^9 x -6.30 x 10^(-4) x 1.07 x 10^(-4)] / (3.00² + 3.00²) = -33.705 N ... [negative just means it's an attractive force]
(d) angle ACB = arctan (3.00 / 3.00) = 45°
so angle F(BC) makes with the positive x-axis is 270 + 45 = 315°
so x-component of F(BC) = -270 sin 315° = 190.9 N ... [using the picture to get the ratio required]
(e) y-component of F(BC) = -270 cos 315° = -190.9 N
(f) x-component of the resultant electric force on C = 0 + 190.9 = 190.9 N
(g) y-component of the resultant electric force on C = 472.5 + -190.9 = 281.6 N
(h) magnitude of the resultant electric force on C = [190.9² + 281.6²] = 340.2 N
Direction of the resultant electric force on C = arctan (190.9 / 281.6) = 34.1° right from the y-axis ... so that is 90 - 34.1 = 55.9° counterclockwise from the +x-axis