Is the Electric Field at Point (-1,0) Correctly Calculated?

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SUMMARY

The electric field at point (-1,0) due to two charges, (-3*10^-6)C at the origin and (6*10^-6)C at (2,0), was calculated to be Etotal = (-2.09*10^4) N/C. The calculation involved summing the contributions from both charges using the formula Etotal = Eq1 + Eq2, where Eq1 = k(q1)/r^2 and Eq2 = k(q2)/(2-r)^2. Despite the negative result, the direction of the electric field indicates it is positive, as it reflects the force on a unit positive charge.

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nyclio
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Two charges located along the x-axis, (-3*10^-6)C at the origin, and (6*10^-6)C at the point (2,0)

What is the electric field at point -1,0?

i did:

Etotal = Eq1+Eq2

=(k(q1)/r^2) + (k(q2)/(2-r)^2)
=(-2.09*10^4) N/C

Can someone please double check my calculations (i did it a few different ways and i always end up with a negative answer)? I plugged this into http://hyperphysics.phy-astr.gsu.edu/hbase/electric/e2p.html#c1 and for some reason their answer comes out positive. Which one is correct?
 
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nyclio said:
Two charges located along the x-axis, (-3*10^-6)C at the origin, and (6*10^-6)C at the point (2,0)

What is the electric field at point -1,0?

i did:

Etotal = Eq1+Eq2

=(k(q1)/r^2) + (k(q2)/(2-r)^2)
=(-2.09*10^4) N/C

Can someone please double check my calculations (i did it a few different ways and i always end up with a negative answer)?
The direction of the field is given by the direction of the force on a unit positive charge (ie. of +1C. ). [itex]\vec E = \vec F/q[/itex]. So, in this case the field is positive.

AM
 

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