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Is the electric field zero?

  • Thread starter BuBbLeS01
  • Start date
602
0
1. Homework Statement
Two point charges are placed along a horizontal axis with the following values and positions: +3.0 �C at x = 0 cm and -7.0 �C at x = 20 cm. At what point along the x axis is the electric field zero?


2. Homework Equations



3. The Attempt at a Solution
Can I just set Ea = Eb
Ea = K*A/R^2
Eb = K*B/R^2
 

Answers and Replies

275
2
thats the basic idea.
but keep in mind that the distances will be different (the R's) and thats what you need to solve for.
Make 2 variables R1 and R2 (for each distance), how are they related?
 
602
0
Oh I thought the distance was just the distance between A and B....but its the distance from point A to the point where the electric field and same for B?
 
275
2
correct. this is a little strange, we're using the method of a "test charge" which essentially means a pretend charge (you can call the charge "q" but it should always end up canceling in cases like this). a test charge is a way to "test" the field at a certain point where there isn't actually a charge. if you use the distance R (from A to B), that will give you the field at A and at B, note that the forces will be the same at those points (in accordance with newton's second law), but you are looking for a point inbetween the A and B where the pulls (or pushes) are equal and cancel eachother out.
 
602
0
So I just use R?
 
275
2
no.
you have to use R1 and R2, where R1 + R2 = R (the total distance)
 
602
0
I don't get it lol...I'm sorry...
 
275
2
Ea = Eb
Ea = K*A/R1^2
Eb = K*B/R2^2
where R2 = R - R1 (in this case R = 20cm the total distance between A and B).
|---------------|-----------------------------|
A_____________C_________________________B
|------R1------||--------------R2------------|
|-------------------R-------------------------|
You're are trying to find the point C, such that the field (or force) is zero.
How about that?
 
602
0
Oh wow thanks :)....I see now....So I end up with...
Ea = K*A/R1^2
Eb = K*B/(R-R1)^2

A/R1^2 = B/(R-R1)^2
And I solve for R1??
 
275
2
exactly!
More importantly, does it makes sense why we set it up like that?
 
602
0
Yup I completely understand why you set it up like that...thanks!!
 

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