E-Fields Question: Finding Zero E-Field & Acceleration

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The discussion focuses on determining the point along the y-axis where the electric field is zero between two fixed charges, +Q and +2Q, located at (0,a) and (0,2a) respectively. It is established that there is indeed a point between the charges where the electric field cancels out, and participants suggest using symmetry and logical reasoning to find this point. For the second part, the acceleration of a small negative charge placed at the origin is calculated using the formula a=Eq/m, with discussions clarifying the correct distances from the charges for the electric field calculations. Participants emphasize the importance of substituting the correct values for the distances from the charges to accurately find the electric field strengths. Overall, the collaborative effort leads to a clearer understanding of the problem and its solution.
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Homework Statement


Two charges, charge one is +Q and charge two +2Q, are fixed in a plane along the y-axis of an x-y coordinate plane. Charge one is at point (0,a) and Charge two is at the point (0,2a).
(a) Is there any portion of the y-axis in which the total electric field is zero? If so, where?
(b) If a small negative charge, -q, of mass m were placed at the origin, determine its initial acceleration (magnitude and direction).

Homework Equations



E1=E2
F=ma
E=F/q
E=1/4pi(epsilon not)xQ/r^2

The Attempt at a Solution



For (a), there will always be an e-field along the y-axis except for a point in between the two charges. I am having trouble finding that point, though. I set the two e-fields equal to each other, but was unsure of what to put as the radius for each.

For (b) I used F=ma and then plugged in F=Eq to get a=Eq/m. My final answer was (-1/4pi(epsilon not)x(Q/a^2+Q/2a^2)xq)/m.
 
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For a) You can use symmetry conditions to find that point. Just think about it logically, where should the E-field go to zero if the 2 charges are equal?

b) Your method is correct. I don't get why you would have an A there since both charges are +Q? Also, the 2a^2 in the denominator of the second term should have parenthesis: (2a)^2
 
For (b) the capital A was supposed to be a Q. And I also mistyped the question--the second charge is +2Q.
 
Ok, then for a)

You have for one charge: E1=kQ/r1^2 and E2=2kQ/r2^2 right. Since you are contained on the y-axis, the distance from a point on the y-axis to your charge 1 and 2 is r1 and r2. That should be relatively easy to find. The variable is y. Find a y such that the two are equal.
 
I think I am still confused on what to substitute for r1 and r2. Should one be y and the other (a-y)?
 
If you are at point y, how far are you from Q1? How far are you from Q2? Those are the values to put for r1 and r2.

If that still confuses you, just put some solid numbers. If I am at y=0, how far am I from Q1 and Q2? What if I were at y=8?
 
So then does y=sqrt(2)a?
 
Thanks a lot for your help! I used real numbers and got the same answer. I understand it much better now.
 
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