• Support PF! Buy your school textbooks, materials and every day products Here!

Finding points where electric potential is 0

  • Thread starter natalie.*
  • Start date
  • #1
8
0
Hi, this is my first time posting on the forums, so if I make any mistakes, I'm sorry.

Homework Statement


This question 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 nC point 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 out.

PART B
b) What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero? (Input 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]

Homework Equations


[tex]V_{net}=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}}[/tex]


The Attempt at a Solution


Simplified down to...
[tex]r_{1}=\frac{q_{1}d}{q_{1}-q_{2}}[/tex]
but this only finds me ONE point where the electric potential is zero. I'm supposed to end up with two...
I can't think of any other way to look at it, or get a quadratic out of it.

Where did I go wrong?
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
Consider the region of the line to the left of the negative charge. The potential due to proximity to the negative charge will be negative. The potential due to proximity to the positive charge will be positive. Now, the positive charge is larger than the negative charge, so you might expect there to be a distance to the left of the negative charge where the magnitudes of their influences will be equal.
 
  • #3
8
0
Yeah, that makes sense to me, because that's how I worked out Part A, but I got the wrong answer for this.
If I work it 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 formula out again, I just get the negative of my first answer, which to me just means my math is saying "You meant to go the other direction"...
If I work out:
[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] originally), my electric potential doesn`t work out to be 0.
 
  • #4
gneill
Mentor
20,793
2,773
What is probably tripping you up is the mixing of assumed directions for radii, and charge signs in the formulae.

When I do these sorts of problems I find it easier to spot the locations where I can expect a zero by eye, then drop 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 give it a try for the point to the left of the negative charge?
 
  • #5
8
0
Awesome, so I know what the two points are now, and I've double checked that the electric potential at both of those points is 0.
I'm still getting the wrong answer though, but I have a few more tries.
Since it asks for the 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 leaving [tex]q_{1}[/tex] with it's negative charge for calculations (since the two charges would exert forces in opposite directions).

Is there anything wrong with that?
 
  • #6
gneill
Mentor
20,793
2,773
Your method looks good. Make sure that when you plug in your numbers for the calculation that you take into account that the units given for d are cm, not m.
 
  • #7
8
0
The directions ended up throwing me off a little, but I finally got it with someone else's help. Thank you!

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]

Note that [tex]r_{2}>r_{1}[/tex], and they are distances from [tex]q_{1}[/tex], which is the negative charge.
 
  • #8
10
0
I'm on this exact problem with different numbers and I'm fairly certain I've done one of the points right (the one to the left of the negative charge) but I can't figure out where the other point is. The math never gives room for a second point, and even when just looking at it there doesn't seem to be a logical second point of zero potential.

Is it at one of the charges? Is it at infinity? My numbers give me a point that's approx. 5cm to the left of q1 and then 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 just comment here so as to not rewrite out the question. Any help would be greatly appreciated.

Nevermind, I re-read the replies and I get it now.
 
Last edited:

Related Threads on Finding points where electric potential is 0

Replies
4
Views
5K
Replies
7
Views
6K
Replies
3
Views
2K
Replies
19
Views
6K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
9
Views
4K
Replies
40
Views
18K
Replies
2
Views
563
Top