What Is the Net Electric Field Midway Between Two Charged Particles?

[Gear300]

Good...thats the magnitude of the net electric field...also, remember that it is pointing to the left (in the negative x-direction); so after adding the magnitudes, put a negative sign in front...so the net electric field is -.1665444.
1. Homework Statement
Two particles are fixed to an x axis: particle 1 of charge -3.50*10-7 C at x = 6.00 cm, and particle 2 of charge +3.50*10-7 C at x = 29.0 cm. Midway between the particles, what is their net ELECTRIC FIELD?

The question is not asking anything for Fe (electric force). Its asking for the net electric field, which you just found.

 

[physicsbhelp]

also astronuc i still don't get what lenghts? the length from each q to 11.5? or is it 11.5+11.5

 

[physicsbhelp]

ooo okay gear300 so now i just plug in -.1665444. for R but if i use the equation e=kq /r^2 then there is only one Q so which q should i use?

 

[Astronuc]

Just follow Gear300.

also astronuc i still don't get what lenghts? the length from each q to 11.5? or is it 11.5+11.5

I think one needs to become familiar with vectors and vector fields. The term length refers to magnitude, rather than simply distance.

Here's a simple tutorial on vectors - http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#veccon

http://hyperphysics.phy-astr.gsu.edu/hbase/vbas.html (Basic concepts)


With respect to the Electric Force and Electric Field, see

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c2

 

[Gear300]

physicsbhelp...we found the answer to the problem...no need to plug anything else in. Thats the answer...there's nothing more...nada...none whatsoever...no continuation...thats the net electric field. E is the net electric field and E = -.1665444 N/C...thats what you were asked to get and you got it. No more worries.
Just do as Astronuc posted just now. Here are a few additional sites:
http://physics.bu.edu/~duffy/PY106/Electricfield.html
and also try looking up superposition in respect to electrical charges and gravitational forces

 

[physicsbhelp]

omg noooOOOOOOOOOOOOOOOOOOOO i typed in the answer that you just said and it was my last chance and we got it WRONG! now i have no more chances!
i got locked out. this was a big part of my grade, but thanks for helping even though we got it wrong.

 

[Gear300]

...oh well...but at least you should have grasped the concept.

 

[Saladsamurai]

Okay. I leave for an hour and all hell breaks lose! You probably need to open your text and do some reviewing here since you seem to be missing the important concepts. This is basic vector addition that is used throughout all of physics.

By drawing the free-body diagram of the electrostatic forces acting on a positive test charge located midway between q1 and q2, one would have realized that both forces are pointing leftward.

By definition, the magnitude of an electric field due to a point charge is E=\frac{k|q_n|}{r^2} and the direction is that of the

electrostatic force.

The net electric field is the summation of all its constituent parts.

E_{net}=\sum_1^nE_n[/itex] <br /> <br /> Since both point leftwards <br /> <br /> E_{net}=-\frac{kq_1}{r^2}-\frac{kq_2}{r^2} where r is in meters<br /> <br /> \Rightarrow E_{net}=-\frac{(8.99\cdot 10^9)(3.5\cdot 10^{-7})}{(11.5\cdot 10^{-2})^2}-\frac{(8.99\cdot 10^9)(3.5\cdot 10^{-7})}{(11.5\cdot 10^{-2})^2}=-476\frac{kN}{C}