Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N
