Question

1. At what separation will the electric force between charges of 2.9 μC and 4.5 μC have a magnitude of 0.35 N ? k=8.99×109N⋅m2/C2.

2. Find the magnitude of the electric force between the charges 0.12 C and 0.33 C at a separation of 2.5 m. k=8.99×109N⋅m2/C2.

3. Three point charges lie on the x axis. Charge 1 (-2.1 μC) is at the origin, charge 2 (+3.2 μC ) is at x = 7.5 cm, and charge 3 (-1.8 μC ) is at x = 11 cm. k=8.99×109N⋅m2/C2. What is the magnitude of the total force exerted on charge 1?

Answer 1

q_1 = 2.9 mu c q_z = 4.5 mu c F = 0.35 N k = 8.99 V 1069 Nm^2/c^2 Let 'd' be the separation b/ wthe q_1 & q_2 F= 1/ 4 pi epsilon_0 q_1 q_2/ d^2 d62 = 8.99 times 10^9 times 2.9 times 4.5 times 10^- 12/ 0.35 d = 0.5789 m = 57/9 cm Q_1 = 0.12 C ,q_2 = 0.33 C d = 2.5 m K = 8.49 times 10^9 Nm^2 /c62 F = 1/ c_1 pi epsilon_0 q_1 q_2/ d62 = 9 times 10^9 times 0.12 times 0.13 /(2.5)^2 = 5.7 times 10^7 N/c F_ net = F_1 - F_ 2 = 1/ 4 pi epsilon_0 q_1 q_2/ (7.5 times 10^- 2) - 1/ 4 pi epsilon_0 q_1 q_3/ (0.11)^2 \r\n F_ nat = 8.99 times 10^9 times 10^- 12 [2.1 times 3.2/(0.075)^2 - 2.1 times 1.8/ (0.11)^2] = 7.93N F_ net = 7.93N Net force on the Chrage 1 is 7.93 N Towards +ve X - axis