A particle with charge 5[tex]\mu[/tex]C is located on the x-axis at the point -6 cm, and a second particle with charge -8 [tex]\mu[/tex]C is placed on the x-axis at 10 cm. The Coulomb constant is 8.9875 x 10^9 Nm^2/C^2. What is the magnitude of the total electrostatic force on a third particle with charge 9[tex]\mu[/tex]C placed on the x-axis at -2 cm. Answer in units of N. First I converted everything. -8[tex]\mu[/tex]C= -8 x 10^-6 C 5 [tex]\mu[/tex]C= 5 x 10^-6 C 9 [tex]\mu[/tex]C= 9 x 10^-6 C -6 cm= -.06 m -2 cm= -.02 m 10 cm=.10 m then I drew a free body diagram for the forces on the third particle and found that both forces point to the right, since the first particle has a positive charge, and the second has a negative charge. Then I used Coulomb's Law, For the first charge, [tex] F= 1/4\pi \epsilon * (5 x 10^-6 * 9 x 10^-6) / (.04)^2 [/tex] Simplifying this gave me [tex] 1/4\pi \epsilon * 2.81 x 10^-8 [/tex] Plugging in 8.9875 x 10^9 for E and solving gave me 2.48 x 10-19. Then I did the same thing for the next charge. [tex] F= 1/4\pi \epsilon * (-8 x 10^-6 * 9 x 10^-6)/ (.12)^2 [/tex] so [tex] 1/4\pi \epsilon * 5 x 10^-9 [/tex] Plugging in for E and solving gave me 4.43 x 10^-20. I then added these together, and got 2.923 x 10^-19 N, which wasn't right. Can someone please help me? I don't understand electrostatics at all!