Question

A charge 5.05 nC is placed at the origin of anxy-coordinate system, and a charge -1.97 nC is placed on the positive x-axis at x = 3.99 cm . A third particle, of charge 6.03 nC is now placed at the point x = 3.99 cm , y = 3.00 cm .

(1)Find the x-component of the total force exerted on the third charge by the other two.

(2)Find the y-component of the total force exerted on the third charge by the other two.

(3)Find the magnitude of the total force acting on the third charge.

(4Find the direction of the total force acting on the third charge.

Answer 1

q1 = 5.05 nC @ x=0
q2 = -1.97 nC @ x = 3.99 cm
q3 = 6.03 nC @ x = 3.99 cm , y = 3.00 cm

Force due to q1 = k q1 q3 / r^2
F1 = 8.9 * 10^9 * 5.05 * 6.03 * 10^-18 / (3.99^2 + 3^2) *10^-4
F1 = 2.71 * 10^-7 / 0.0025
F1 = 1.1 * 10^-4 N/C

cos(theta) = 3/ 5
sin(theta) = 3.99 / 5

F1 x = F1 cos(theta) = 1.1 * 10^-4 * (3/5)
F1 x = 6.6 * 10^-5 N/C

F1 y = F1 sin(theta) = 1.1 * 10^-4 * (3.99/5)
F1 y = 8.8 * 10^-5 N/C

Force due to q2 = k q2 q3 / r^2
F2 = 8.9 * 10^9 * 5.05 * 1.97 * 10^-18 / (0.03^2)
F2 = 8.854 * 10^-8 / 0.0009
F2y = 9.8 * 10^-5 N/C

Fx = F1x
Fx = 6.6 * 10^-5 N/C
x-component of the total force exerted Fx = 6.6 * 10^-5 N/C

Fy = F1y - F2y
Fy = (8.8 - 9.8 ) * 10^-5 N/C
Fy = - 1 * 10^-5 N/C (negative sign indicates direction as -ve Yaxis )
y-component of the total force exerted Fy = - 1 * 10^-5 N/C

F = sqrt(Fx^2 + Fy^2)
F = sqrt (( 6.6 * 10^-5)^2 + ( 1* 10^-5)^2)
F = 6.68 * 10^-5 N
Magnitude of the total force acting on the third charge F = 6.68 * 10^-5 N

theta = tan^-1(6.6 /1)
theta = 81.38⁰

theta = 8.62⁰ from +ve X axis in Clockwise Direction.