Finding points where electric potential is 0

Click For Summary
SUMMARY

This discussion centers on calculating the electric potential and electric field at points along the x-axis between a -10.1 nC point charge and an +18.9 nC point charge, which are 13.8 cm apart. The user successfully determined one point where the electric potential is zero but struggled to find a second point. Key equations used include V_{net}=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}} and E=|\frac{Kq_{1}}{r_{1}^{2}}+\frac{Kq_{2}}{r_{2}^{2}}|. The discussion concludes with the realization that the second point of zero potential exists to the left of the negative charge, where the influences of both charges balance out.

PREREQUISITES
  • Understanding of electric potential and electric field concepts
  • Familiarity with Coulomb's Law and point charge interactions
  • Ability to manipulate algebraic equations involving distances and charge magnitudes
  • Knowledge of units conversion, specifically between centimeters and meters
NEXT STEPS
  • Study the principles of electric potential and electric field in electrostatics
  • Learn how to apply Coulomb's Law in various configurations of point charges
  • Explore the concept of equipotential surfaces and their relationship to electric fields
  • Investigate the mathematical techniques for solving quadratic equations in physics problems
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields and potentials in multi-charge systems.

natalie.*
Messages
6
Reaction score
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)

q_{1}=-10.1 \times 10^{-9} C

q_{2}=18.9 \times 10^{-9} C

d=0.138 m

r_{2}=d-r_{1}

r_{1}=r_{1}

Homework Equations


V_{net}=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}}


The Attempt at a Solution


Simplified down to...
r_{1}=\frac{q_{1}d}{q_{1}-q_{2}}
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?
 
Physics news on Phys.org
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.
 
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 r_{1} is to the left of q_{1}, then r_{2}=d+r_{1}.
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:
V=\frac{Kq_{1}}{r_{1}}+\frac{Kq_{2}}{r_{2}}
where I set r_{2}=d+r_{1} (using the same positive value I found for r_{1} originally), my electric potential doesn`t work out to be 0.
 
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:

\frac{|q1|}{r} = \frac{|q2|}{d - r}}

Why not give it a try for the point to the left of the negative charge?
 
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:
E=|\frac{Kq_{1}}{r_{1}^{2}}+\frac{Kq_{2}}{r_{2}^{2}}|
where K=8.99\times10^{9}
and I'm leaving q_{1} with it's negative charge for calculations (since the two charges would exert forces in opposite directions).

Is there anything wrong with that?
 
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.
 
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:
E_{1}=\frac{K|q_{1}|}{r_{1}^{2}}+\frac{K|q_{2}|}{(d-r_{1})^{2}}
Answer 2:
E_{2}=\frac{K|q_{1}|}{r_{2}^{2}}-\frac{K|q_{2}|}{(d+r_{2})^{2}}

Note that r_{2}>r_{1}, and they are distances from q_{1}, which is the negative charge.
 
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:

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
64
Views
5K
  • · Replies 7 ·
Replies
7
Views
890
Replies
1
Views
821
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
673
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K