Electric Field with three negative point charges in a line

AI Thread Summary
The discussion focuses on calculating the electric field at point P due to three negative point charges arranged in a line. The user correctly identifies that the net electric field is the vector sum of the fields from each charge. They calculated the electric fields for the -5μC charges as -4,500,000 N/C each and for the -2μC charge as -5,000,000 N/C. However, they were reminded that the electric field is a vector quantity, necessitating consideration of direction in their calculations. The user ultimately seeks clarification on their approach and whether they made any errors in their calculations.
kddc
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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.
 
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kddc said:

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.
 
Thank you!


Forgot to log in after having solved it.
 
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