Net electric field

1. Jan 31, 2017

Jrlinton

1. The problem statement, all variables and given/known data

2. Relevant equations
Enetx=((kq1)/r^2 +^2) with the pos/neg of the coefficients being dependent on the position of the charges around the point

3. The attempt at a solution
As everything is one dimensional we can assume the j an k compnents of the electric field are 0
Part a
Ea= -(kq1)/r1^2+(kq2)/r2^2
=-(8.99E9N)(3.7E-12C)/(.011m)^2+(8.99E9N)(-1.8E-12C)/(.022m)^2
=-274.901 N/C+ -33.4339 N/C
=-308.335 N/C

This was incorrect and I am unsure of my mistake(s) and can only assume I would carry them over to the next two calculations

2. Jan 31, 2017

kuruman

At A, the positive charge provides a negative component to the left and the negative charge provides a positive component to the right. You show two negative components. In short, you used the negative sign for the negative charge twice. It is probably easier to calculate the size of the component using positive values for all charges, figure out the direction of the field components from the diagram and add negative signs as needed.

3. Jan 31, 2017

Jrlinton

So you're saying I should use absolute values for both charges and hen decide direction when adding them together? So in this case it should have been -274.901+33.4339= -241.467 N/C?

4. Jan 31, 2017

Staff: Mentor

The signs of the charges along with the relative locations of the charges at the points of interest are mucking up your algebra. You either have to be very careful indeed to keep it all straight, or do the practical thing: First sketch in vectors for the field directions due to each charge at the points of interest. Then calculate the field magnitudes using absolute values for the charges. Write the algebra to incorporate direction based on the vector directions.

Edit: Ah. I see that kuruman got there ahead of me!

5. Jan 31, 2017

kuruman

That's what I and @gneill are saying. Now you should be able to do the other two parts on your own. If not, you know where to go for help.