Question

12.4 pC Pictured on the right are three point charges Q- 12.4 JC, 02-34.6 pC, and Qs 57.3 C arranged according to the figure on the right. A fourth point charge is located at point A with a charge of Q-17.5 C. Calculate the magnitude of the net force on the charge at point A 17.5 uC 80.1 cm Number -34.6 C o, 157.3 LC k-一80.1 cm

Answer 1

As you know the same charge repels each other and opposite charge attracts each other.

Here Q_1 = 12.4 mu c Q_2 = 34.6 mu c Q_3 = 57.3 mu c Q_A = 17.5 mu c Here Q = 45 degree x-component of the force F_x = F_1 - F_2 sin Q = k Q_1 Q_A/r^2_1 - k Q_1 Q_A/r^2 sin Q Here r_1 = 0.801 m r = squareroot 0.801^2 + 0.801^2 r = 1.133 m k = 3 times 10^9 unit = > F-x = 9 times 10^2 times 12.4 times 10^-6 times 17.5 times 10^-6/(0.801)^2 - 9 times 10^9 times 34.6 times 10^-6 times 17.5 times 10^-6 times sin/(1.133)^2 = 3.044 - 3.002 = 0.042 F_x = 0.042 N y - component of the force F_y = F_3 - F_2 cos Q = > F_y = k Q_3 Q_A/r^2_3 - k Q_2 Q_A/r^2 cos Q = 9 times 10^9 times 57.3 times 10^-6 times 17.5 times 10^-6/(0.801)^2 - 9 times 10^9 times 34.6 times 10^-6 times 17.5 times 10^-6 cos (45)/(1.133)^2 = 14.066 - 3.002 = 11.064 N So Resultant force on the