Calculating the magnitude of point charges

[ezperkins]

Homework Statement

A point charge of -6 µC is located at x = 1 m, y = -2 m. A second point charge of 12 µC is located at x = 1 m, y = 3 m.

(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0.


(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0.

Homework Equations

The Attempt at a Solution

$$\Sigma_{F}= k\frac{-6 \mu C}{8} + k\frac{12 \mu C}{13} = 1.08e4 \frac{N}{C}$$


That's for the magnitude part of A. I've already submitted to many incorrect responses, but I would like to know how to do it. Thanks ahead of time.

 

[hage567]

Homework Statement

A point charge of -6 µC is located at x = 1 m, y = -2 m. A second point charge of 12 µC is located at x = 1 m, y = 3 m.

(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0.


(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0.

Homework Equations

The Attempt at a Solution

$$\Sigma_{F}= k\frac{-6 \mu C}{8} + k\frac{12 \mu C}{13} = 1.08e4 \frac{N}{C}$$


That's for the magnitude part of A. I've already submitted to many incorrect responses, but I would like to know how to do it. Thanks ahead of time.

You need to find both the x and y components from each charge at the point of interest and add them up accordingly to find the resultant field.

 

[ezperkins]

I did that for another problem that asked for the respective components, why is it necessary that I so that for this problem?

 

[hage567]

Since I don't know what the other question was, I can't tell you what's different. You need to do vector components because the point of interest is not on the same line as the two charges. It is a two dimensional situation. Have you tried what I suggested?