Electric Field with three negative point charges in a line

[kddc]

Homework Statement

Find the magnitude of the electric field this comination of charges produces at point P, which lies 6.00cm from the -2.00μC charge measured perpendicular to the line connecting the three charges.

Homework Equations

E=k$\frac{Q}{r^{2}}$
F=k$\frac{q_{1}q_{2}}{r^{2}}$
F=$\frac{E}{q}$
k$\approx9*10^{9}$

The Attempt at a Solution

Conversions to the correct units were done prior to plugging values into eqn.

To my understanding the net electric field is the sum of all electric fields from each point charge.
So I found E$_{-5μC}$=-4,500,000 and the same goes for the other -5μC point charge; E$_{-5μC}$=-4,500,000 , and for the -2μC point charge I got E$_{-2.00μC}$= -5,000,000

For E$_{-5μC}$ I calculated and then plugged in $\sqrt{(8cm^{2}+6cm^{2}}$≈0.1m


Results summed up to E$_{-5μC}$+E$_{-5μC}$+E$_{-2.00μC}$=(-4.5*10$^{6}$)+(-4.5*10$^{6}$)+(-5.0*10$^{6}$)= -1.40*10$^{7}$

The question only asked for magnitude so I'm assuming direction doesn't really matter at this point. What am I doing wrong? Please help.

 

[tms]

Homework Statement

Find the magnitude of the electric field this comination of charges produces at point P, which lies 6.00cm from the -2.00μC charge measured perpendicular to the line connecting the three charges.

Homework Equations

E=k$\frac{Q}{r^{2}}$
F=k$\frac{q_{1}q_{2}}{r^{2}}$
F=$\frac{E}{q}$
k$\approx9*10^{9}$

The Attempt at a Solution

Conversions to the correct units were done prior to plugging values into eqn.

To my understanding the net electric field is the sum of all electric fields from each point charge.

So far so good.

So I found E$_{-5μC}$=-4,500,000 and the same goes for the other -5μC point charge; E$_{-5μC}$=-4,500,000 , and for the -2μC point charge I got E$_{-2.00μC}$= -5,000,000

The electric field is a vector, and you must take the vector sum of the fields of the three charges.

 

[kddc]

Thank you!


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