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

[physicsbhelp]

omg usually i an soo careful with units and i though i even checked my units this time but i didnt convert cm to meters!
okay so this is what i got for my answer -0.0095790

NOW THE question is if it should be negative or not because even though i got a negative answer the force would be positive rihgt?

 

[physicsbhelp]

BUT I JUST TYPED IN THE ANSWER AND IT SAYS I GOT IT WRONG again!

i typed in the negative with it so is that why it is wrong?

this ismy last chance for the correct answer beucase i answer wrong again i will be locked out from the question.

 

[Saladsamurai]

okay so this is what i got for E for Q1 :-23.798858

Why is E1 negative, but E2 is positive? When you calculate E, the direction is given by the direction of the electrostatic force between the particle and the TEST charge. Again, from the free-body diagram (which I am so persistent to try and get people to actually use!) we can see that the directions of F1 and F2 are the SAME. Thus, E1 and E2 must be the same as well.

 

[Saladsamurai]

Also. Looking at post#37. IF you did have the sign correct (which you do not) why would they add to a number? If they are equal and opposite (which you said they were in #29 and #30, they should have summed to zero!

 

[physicsbhelp]

okay okay so suppose they are not negative, it wouldn't make much of a difference exept for the sign.
but what did i do wrong?? :( i can't believe i got the wrong answer.

 

[Saladsamurai]

You are not thinking clearly. What is the difference between -A+A and -A-A ?

One equals___and the other equal -_____

Fill in the blanks and see your error.

 

[physicsbhelp]

Wait i am not so sure of what you are saying now. what do you mean that i added the numbers? which numbers?
i just thought i had to plug in the given information into the equation. how am i supposed to get numbers from the free-body diagram? like i just don't understand. R=11.5 and k=8.99E9 and q1=-3.50e-7 and q2=3.50e-7 so why sould i add q1 and q2 together?

 

[physicsbhelp]

i get what you are saying that for example -6+6=0 but why would i want to add q1 and q2 together? i mean what would that give me, and where would i put that in the equation

 

[Saladsamurai]

You need to go back over what I have told you. Find the post on the principle of superposition. Post #10.

 

[physicsbhelp]

i found the post and read it. but i don't understand what should be 0 because if i put in 0 for Q in the Equation E=KQ /R^2 then E would equal 0 so I am not quite sure what should be 0 in that equation?

 

[Saladsamurai]

0 is a point (the mid point) not a number.

 

[Gear300]

physicsbhelp, every charge produces its own electric field. An electric field is basically how a charge electrically influences its surrounding environment and is equal to E=KQ /R^2. In this case, you have 2 electric fields because you have 2 charges. The question is basically what the net electric field is at the midpoint. Keep in mind that electric field is a vector. The direction it has is either away from the charge or towards the charge. For negative charges, the direction is towards the charge, and for positive charges, the direction is away from the charge. One of your charges is negative and one is positive. So here's what you do, draw the 2 charges and put a dot for the midpoint. Then, figure out the direction of each electric field at the midpoint and draw and arrow to represent it.

 

[Saladsamurai]

Find E at point 0 due to particle 1. Then find E at O due to particle 2. Then add them algebraically. I am not sure how else to say it. You have all you need.

Goodnight.

 

[physicsbhelp]

wait please don't leave!
so what i do is E= k (-3.501e-7) / (+.06)
and E = (8.99e9)(3.50e-7) / (-.29)

 

[Gear300]

Check if you did it my way...it'll make it easier for me to explain. Once you've done the diagram I described, what direction do you notice the two arrows are going?

 

[physicsbhelp]

yes gear300 i had already made a vecotr diagram i think it is on page 1 or 2 of this thread. can you look at it

 

[physicsbhelp]

okay gear300: is this what you mean: F01<-----QO<-----F02

where F01 is the "force on 0 ue to 1" and
F02 is the force on 0 due to 2.

As you can see, i have drawn F01 as attracting and F02 as repelling

 

[Gear300]

No...remember, an electric field is different from electric force. There are 2 separate electric fields: one of them is produced by the charge at 6.00cm and the other at 29.00cm. Right now, use the equation you have for electric field to find the electric field for each charge at the midpoint. Remember that R for both of them is 11.5cm and also remember to convert to meters.

 

[physicsbhelp]

Okay so the E1=-874027.778
and E2=37413.7931 OR would it be -37413.7931 because from Q2 to 0 is -.29meters

 

[physicsbhelp]

NOW GEAR300 AND SALADSAMURAI are CONTRADICTING EACH OTHER! because saladsamurai is saying that the R for the electric field equation is the distnace from q1 to 0 and from q2 to 0 which is 6 and 29 while GEAR300 is saying that it is 11.5


I AM SOO CONFUSED





SRY!

 

[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 the 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]

because 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 the 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?