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

[Gear300]

No...those look kind of wrong.
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?

Electric field is E= KQ/R^2. There are 2 electric fields because there are 2 charges.

E1 = K(+3.50*10E-7C)/(.115m)
E2 = K(-3.50*10E-7C)/(.115m)...This is because R for both of them are is 11.5cm, which converts to .115m
Now plug in the numbers in a calculator and solve for E1 and E2.

 

[physicsbhelp]

okay q1 is negative you said it was positive accidentally but whatever

okay so E1= K(-3.50*10E-7C)/(.115m) which = -0.0832722117
E2 = K(3.50*10E-7C)/(.115m) Which = 0.0832722117

 

[Gear300]

One of them have a negative sign. What that means is that whatever distance R you are from that charge, the electric field is pointing towards that charge. If the sign is positive, then the electric field is point away from the charge. The charge with the negative electric field (E1), is -3.50*10E-7C and is located at 6.00cm. The charge with the positive electric field (E2) is 3.50*10E-7C and is located at 29.0cm.

<_______q1_______MP______q2______> [x-axis]

Alright, q1 is the negative charge with the negative electric field and q2 is the positive charge with the positive electric field. MP is the midpoint. Mentally picture arrows at MP, in which each arrow is the direction its respective electric field is taking (you'll have 2 arrows because you have 2 electric fields). Remember: If the electric field is negative, it is pointing towards that charge. If the sign is positive, then the electric field is point away from the charge. Now, what direction, to the left or right, are both arrows pointing?

 

[physicsbhelp]

both arrows are pointing to the left right?

 

[Gear300]

Yes...both are pointing to the left.

So here is what you do now.
E1 = -0.0832722117 and E2 = 0.0832722117
Ignore the sign on E1...that just means that the electric field of E1 is pointing towards its respective charge. Just add the Magnitudes of both electric fields to get the net electric field. Now you also know that the electric field is to the left, which is also in the negative x-direction...so the net electric field you calculated is pointing in the negative x-direction.

 

[physicsbhelp]

okay thanks!
but what are teh magnitudes? like are they the E's?

 

[Astronuc]

The problem description states that "Two particles are fixed to an x axis", so I am not sure why the pictures show two axis, one horizontal and one vertical.

q1--------------0--------------q2

and one is asked to find the E field at the midpoint, as saladsamurai indicated in post #8.

There is one electric field which is composed of the electric (vector) fields of both charges.

Using F=Eq, then E = F/q (see post #13, where one is correct), and one may assume a positive test (reference) charge, so if

F=kq_iq/r² then E_i= kq_i/r²


By convention, a negative charge has the electric field vector pointing toward it, while the positive charge has the electric field pointing away from it.

Remember to square the distance between the charges and the point of interest. One must also use the appropriate unit vector to indicate the direction of the charge.

 

[physicsbhelp]

becuase if i add the E's together i get .1665444 when i ignore the negative sign.
so now do i plug this into Fe=...equation?

 

[Astronuc]

okay thanks!
but what are teh magnitudes? like are they the E's?

In a vector, the magnitude is the 'length' and is positive. The - and + in the electric field indicated direction with respect to some reference coordinate system, more or less as Gear300 indicated.

 

[physicsbhelp]

OMG THANK YOU ASTRONUC buT what i still don't understand is what would R be in the equation Ei= kq1/r2 would it be 11.5?

 

[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...theres 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}