Electric field between -8.0 µC and +6.0 µC charges 4 cm apart

[joej]

I'm doing the following problem, it seems to me that I am doing all the correct steps but my answer differs from the answer key, am I missing something?

1. What is the magnitude and the direction of the electric field at a point midway in between a -8.0uC and a +6.0uC change 4cm apart.


This is what I'm doing:

E = ( k * (Q1 / r1^2) ) * ( 1 + ( (Q2 / Q1) / (r2^2 / r1^2) ) )

==

(9x10^9 * ( (8 * 10^-6) / ( 2 * 10^-2) ) ) * ( 1 + ( (6 / 8) / (2 / 2) ) )

==

3600000 * 1.75 = 6300000 N/C


now... that is what I'm getting, could somoen please point out to me what I am forgetting to do, if anything.



__________edit

stupidity strikes again... forgot to square r1

 

[Galileo]

Why write E in such a strange way? (to me anyway)

Use Coulomb's law and the superposition principle.
Calculate the field due to Q1 and the field due to Q2, then add.

 

[maverick280857]

Hello joej...

The formula for electric field at a point distant r units from a charge Q is:

$$\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}\hat{e}_{r}$$

The direction of the field is radial. In your problem you need to compute

$$\vec{E}_{net} = \vec{E_{1}} + \vec{E_{2}}$$

the two terms arising due to the electric fields produced (independently--make a note of this, the superposition principle) has been used here) by the two charges. Have you used this fact?

Cheers
Vivek