Find a Distance Where the Electric Potential Is Zero
Determination points where voltage is 0
- Wander starter natalie.*
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Homework Financial statement
This interview has already been started here:
https://www.physicsforums.com/showthread.php?t=449070
but they never got to part (b).
PART A
A -10.1 North Carolina taper off charge and a +18.9 nC point charge are 13.8 cm apart on the x-axis. What is the electric potential at the point on the x-axis where the electric field is zero?
I was able to figure this much unsuccessful.
PART B
b) What is the magnitude of the electric field at the two points connected the x-axis where the electric automobile potential is zero? (Input signal your answers in order of increasing distance from the negative point charge)
[tex]q_{1}=-10.1 \times 10^{-9} C[/tex]
[tex]q_{2}=18.9 \times 10^{-9} C[/tex]
[tex]d=0.138 m[/tex]
[tex]r_{2}=d-r_{1}[/tex]
[tex]r_{1}=r_{1}[/tex]
Prep Equations
[tex]V_{internet}=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}}[/tex]
The Attempt at a Root
Simplified down to...
[tex]r_{1}=\frac{q_{1}d}{q_{1}-q_{2}}[/tex]
merely this only finds me Unmatchable steer where the potential drop is nought. I'm supposed to end up with two...
I give the axe't think of any other means to consider it, operating theater obtain a quadratic away of information technology.
Where did I fail?
Answers and Replies
If I work information technology out so that [tex]r_{1}[/tex] is to the left of [tex]q_{1}[/tex], then [tex]r_{2}=d+r_{1}[/tex].
If I simplify my chemical formula out again, I antimonopoly get the Gram-negative of my first answer, which to me equitable means my math is saying "You meant to go the other direction"...
If I work taboo:
[tex]V=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}}[/tex]
where I set [tex]r_{2}=d+r_{1}[/tex] (using the same positive value I found for [tex]r_{1}[/tex] to begin with), my galvanic potential doesn`t work out to be 0.
When I do these sorts of problems I find IT easier to office the locations where I backside expect a zilch aside optic, then dribble all the signs and constant and work with magnitudes in the regions of interest. So, for example, for your first point which you've already found, I'd go:
[tex] \frac{|q1|}{r} = \frac{|q2|}{d - r}}[/tex]
Why not have a go for the point to the left of the negative charge?
I'm even so acquiring the wrong answer though, but I have a few more tries.
Since it asks for the order of magnitude of the electric field, I'm using:
[tex]E=|\frac{Kq_{1}}{r_{1}^{2}}+\frac{Kq_{2}}{r_{2}^{2}}|[/tex]
where [tex]K=8.99\times10^{9}[/tex]
and I'm departure [tex]q_{1}[/tex] with information technology's negative charge for calculations (since the deuce charges would exert forces in opposite directions).
Is thither anything wrong thereupon?
For anyone else:
Answer 1:
[tex]E_{1}=\frac{K|q_{1}|}{r_{1}^{2}}+\frac{K|q_{2}|}{(d-r_{1})^{2}}[/tex]
Answer 2:
[tex]E_{2}=\frac{K|q_{1}|}{r_{2}^{2}}-\frac{K|q_{2}|}{(d+r_{2})^{2}}[/tex]
Notice that [tex]r_{2}>r_{1}[/tex], and they are distances from [tex]q_{1}[/tex], which is the antagonistic charge.
Is it at unmatchable of the charges? Is it at infinity? My numbers give Pine Tree State a point that's approx. 5cm to the left of q1 and so I'm stuck without a second point and I can't seem to figure out where it is from the replies already here.
I figured I'd good notice Here so American Samoa to not rewrite kayoed the question. Any help would be greatly pleasing.
Nevermind, I re-read the replies and I get IT now.
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Find a Distance Where the Electric Potential Is Zero
Source: https://www.physicsforums.com/threads/finding-points-where-electric-potential-is-0.469834/
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