(i) Part 1
Net force on 3 uC charge
Force due to - 6 uC
[tex]F_1 = \frac{(9\times 10^9)(3\times 10^{-6})(6\times 10^{-6})}{0.15^2}[/tex]
[tex]F_1 = 7.2 N[/tex]
Force due to -2.4 uC charge
[tex]F_2 = \frac{(9\times 10^9)(3\times 10^{-6})(2.4\times 10^{-6})}{0.15^2}[/tex]
[tex]F_2= 2.88 N[/tex]
Force due to 9 uC charge
[tex]F_3 = \frac{(9\times 10^9)(3\times 10^{-6})(9\times 10^{-6})}{2(0.15)^2}[/tex]
[tex]F_3 = 5.4 N[/tex]
Now net force is given as
[tex]F_x = F_1 + F_3cos45[/tex]
[tex]F_x = 11.02 N[/tex]
[tex]F_y = F_2 + F_3sin45[/tex]
[tex]F_y = 6.70 N[/tex]
[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F_{net} = 12.9 N[/tex]
Part 2
Net force on -6 uC charge
Force due to 3 uC
[tex]F_1 = \frac{(9\times 10^9)(3\times 10^{-6})(6\times 10^{-6})}{0.15^2}[/tex]
[tex]F_1 = 7.2 N[/tex]
Force due to 9 uC charge
[tex]F_2 = \frac{(9\times 10^9)(6\times 10^{-6})(9\times 10^{-6})}{0.15^2}[/tex]
[tex]F_2= 21.6 N[/tex]
Force due to -2.4 uC charge
[tex]F_3 = \frac{(9\times 10^9)(6\times 10^{-6})(2.4\times 10^{-6})}{2(0.15)^2}[/tex]
[tex]F_3 = 2.88 N[/tex]
Now net force is given as
[tex]F_x = F_1 - F_3cos45[/tex]
[tex]F_x = 5.16 N[/tex]
[tex]F_y = F_2 + F_3sin45[/tex]
[tex]F_y = 23.6 N[/tex]
[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F_{net} = 24.2 N[/tex]
PART 3
Net force on 9 uC charge
Force due to - 6 uC
[tex]F_1 = \frac{(9\times 10^9)(9\times 10^{-6})(6\times 10^{-6})}{0.15^2}[/tex]
[tex]F_1 = 21.6 N[/tex]
Force due to -2.4 uC charge
[tex]F_2 = \frac{(9\times 10^9)(9\times 10^{-6})(2.4\times 10^{-6})}{0.15^2}[/tex]
[tex]F_2= 8.64 N[/tex]
Force due to 3 uC charge
[tex]F_3 = \frac{(9\times 10^9)(3\times 10^{-6})(9\times 10^{-6})}{2(0.15)^2}[/tex]
[tex]F_3 = 5.4 N[/tex]
Now net force is given as
[tex]F_x = F_2 - F_3cos45[/tex]
[tex]F_x = 4.82 N[/tex]
[tex]F_y = F_1 - F_3sin45[/tex]
[tex]F_y = 17.8 N[/tex]
[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F_{net} = 18.4 N[/tex]
