Particle A of charge 3.15 10 4 C is at the origin, particle B of charge 6.15 10
ID: 1591721 • Letter: P
Question
Particle A of charge 3.15 104 C is at the origin, particle B of charge 6.15 104 C is at (3.76 m, 0) and particle C of charge 1.30 104 C is at (0, 3.70 m).
(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 to obtain the resultant x-component of the electric force acting on C.
(h) Find the magnitude and direction of the resultant electric force acting on C.
Explanation / Answer
Draw the picture
(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 3.15 x 10^(-4) x 1.30 x 10^(-4)] / 3.70² = 269.2 N
(c) magnitude of F(BC) = [9 x 10^9 x -6.15 x 10^(-4) x 1.30 x 10^(-4)] / (3.70² + 3.70²) = -262 N ... [negative just means it's an attractive force]
(d) angle ACB = arctan (3.70 / 3.70) = 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