Electric Field Done Correctly?

[nyclio]

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?

 

[Andrew Mason]

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. ). $\vec E = \vec F/q$. So, in this case the field is positive.

AM

 

[kyle8921]

I cannot confirm the accuracy of your calculations without seeing your work and knowing the specific values of the constants you used. However, I can provide some general feedback on your approach.

Firstly, it is important to note that the electric field at a point is a vector quantity, meaning it has both magnitude and direction. Therefore, it is important to include the direction in your answer, not just the magnitude.

Secondly, when calculating the electric field due to point charges, it is important to consider the direction of the electric field due to each charge separately, and then add them together using vector addition.

In this case, the first charge (q1) is located at the origin, which means the electric field at point (-1,0) will be directed towards the left. The second charge (q2) is located at (2,0), which means the electric field at point (-1,0) will be directed towards the right. Therefore, when adding the electric field due to these two charges, you should end up with a net electric field directed towards the right.

I cannot confirm the accuracy of the answer provided by the website you mentioned, but based on the information you provided, it seems like their answer may be more accurate since it is positive and directed towards the right.

In conclusion, while your approach may be correct, it is important to carefully consider the direction of the electric field and use vector addition to accurately determine the net electric field at a point. I would suggest double checking your calculations and seeking help from a teacher or instructor if needed.