Question

5) At the corners of an equilateral triangle, side L = 0.2 m, there are three charged particles, q1 = 7.00 μC, q2 = 3 μC and q3 = -4.00 μC as seen in the figure. Calculate the total electric force on the load of value 7.00 μC.

9 60.0 92 9s

Answer 1

Electric force due to a set of curves can be found by first finding force due to one charge on other individually and then finding their resultant using rules of vector addition.

Calculation and work is shown below.

Given L = 0.2 m F_12 = Force on charge q_1 due to q_2 F_13: Force on charge q_1 due to q_3 and is attractive unlike charges attract Electric force is given by, F = 1/4 pi epsilon_0 q_1 q_2/r^2 1/4 pi epsilon_0 = constt = 9 times 10^9 N - m^2/c^2 r = separation b/w charges F_12 = 1/4pi epsilon_0 (7 mu c) (3 mu c)/(0.2)^2 m^2 = 9 times 10^9 N - m^2/c^2 times 7 times 10^-6 times 3 times 10^-6/0.04 m^2 F_12 = 4.725N F_13 = 1/4 pi epsilon_0 (7 mu c) (4 mu c)/(0.2)^2 m^2 = 9 times 10^9 N - m^2/c^2 = 9 times 10^9 N - m^2/c^2 times 7 times 10^-6 times 4 times 10^-6 c^2/0.04 m^2 F_13 = 6.3 N Total force = F resultant = squareroot F^2_12 + F^2_13 + 2 times F_12 times F_13 times cos 120 degree F_R = squareroot (4.725)^2 + (6.3)^2 + 2 times 4.725 times 6.3 times cos (120 degree) F_R = 5.679 N So, Total force = 5.679 N