I have 3 point charges placed at the x-axis in the table below I will show their positions. \begin{array}{|c|c|c|c|}\hline\mathrm{q_1}&2\ \mathrm{\mu C}&x_1&0\ \mathrm{m}\\ \hline \mathrm{q_2} & -3\ \mathrm{\mu C}& x_2 & .40\ \mathrm{m}\\ \hline \mathrm{q_3} &-5\ \mathrm{\mu C} & x_3 & 1.20\ \mathrm{m} \\ \hline \end{array}
These points are reference points from the origin meaning its $.4$ meters from the origin and so on. My question is why is my summation wrong I calculate summation of forces acting on the second point charge, $q_2$. This is how I did it: \begin{equation}\hat{F}=k\frac{qq}{r^2}\hat{r} \end{equation} I also used $k$, Coulomb's Constant, to be defined as $9.00\cdot 10^9 \mathrm{\frac{N\cdot m^2}{C^2}}$ So for my attempt I did the following: \begin{align}\sum F&=F_{21}+F_{23}\\ &=(9.00\cdot10^9)(\frac{-3\cdot 10^{-6}\times 2\cdot 10^{-6}}{.4^2}+\frac{-5\cdot 10^{-6}\times -3\cdot10^{-6}}{.8^2}) \\ &=-0.127\ \mathrm{N} \end{align} I have the Schaum's 3000 problem Physics Series and it says that the answer is $-.55\ \mathrm{N}$. Am I wrong?
Edit
I would like to point out that my subscript notation was also wrong here, and here is the corrected one. \begin{align}\sum F&=F_{12}+F_{32}\\ &=(9.00\cdot10^9)(\frac{-3\cdot 10^{-6}\times 2\cdot 10^{-6}}{.4^2}+\frac{-5\cdot 10^{-6}\times -3\cdot10^{-6}}{.8^2}) \\ &=-0.127\ \mathrm{N} \end{align}
After realizing my mistakes, and taking into account the repulsive force that Charge 2 experinces is the opposite direction of the positive x. So taking that into account I would need to add a negative sign in my solution to get the right solution to the problem. \begin{align}\sum F&=F_{21}+F_{23}\\ &=(9.00\cdot10^9)(\frac{-3\cdot 10^{-6}\times 2\cdot 10^{-6}}{.4^2}+-\frac{-5\cdot 10^{-6}\times -3\cdot10^{-6}}{.8^2}) \\ &=-0.550\ \mathrm{N} \end{align}
Edit
I would like to admit that the subscript notation is also wrong in this answer. \begin{align}\sum F&=F_{12}+F_{32}\\ &=(9.00\cdot10^9)(\frac{-3\cdot 10^{-6}\times 2\cdot 10^{-6}}{.4^2}+-\frac{-5\cdot 10^{-6}\times -3\cdot10^{-6}}{.8^2}) \\ &=-0.550\ \mathrm{N} \end{align}