What is the Electric Potential at Points on the Y-Axis with Zero Potential?

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Homework Help Overview

The problem involves determining the points on the y-axis where the electric potential is zero due to two point charges located on the x-axis. The charges are -3.0 nC at x = -9 cm and +4.0 nC at x = 16 cm.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the electric potential formula and the need to consider distances from both charges to points on the y-axis. There are questions about the appropriateness of selecting arbitrary points on the y-axis and how to set up the equations correctly.

Discussion Status

Participants are actively engaging with the problem, attempting to derive the necessary equations and questioning their understanding of the setup. Some guidance has been provided regarding the use of the Pythagorean theorem to find distances, and there is an ongoing exploration of how to manipulate the equations to find the variable y.

Contextual Notes

There is a noted complexity in solving the equations due to the presence of multiple unknowns, and participants are encouraged to plug in specific values for the charges to progress further.

Mitchtwitchita
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Homework Statement



A -3.0 nC charge is on the x-axis at x = -9cm and a +4.0 nC charge is on the x-axis at x = 16 cm. At what point or points on the y-axis is the electric potential zero?


Homework Equations



V = kq/r

The Attempt at a Solution



I know the answer is +/- 12 cm but I have no clue as to how that number was achieved. Can anybody please help me out with this one?
 
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The potential at a point at distance r from a charge q is kq/r. If there are more charges, their potentials add up. You have two charges placed on the x axis. Take on a point on the y axis, and write out the distance from both charges. It is easier if you make a drawing. Use this distances as "r" in the formula for the potential. Add up both terms and take it equal to zero. Solve the equation for y.

ehild
 
So, I can take any arbitrary point on the y-axis, and it will eventually lead me to the right answer?
 
Yes, if your procedure is correct. Let me see how you proceed.

ehild
 
So, I took an arbitrary point of (0, 5 cm), and this is what I got:

r1 = 1.03 x 10^-1 m
r2 = 1.68 x 10^-1 m

Vi = kq1/r1 + kq2/r2
= 2.62 x 10^2 V/m + 2.14 V/m
= 4.76 x 10^2 V/m

Now, I don't know where to go from here.
 
Arbitrary means just y. Do not give any special value. Write the formula for the distance of the charge at point (-9,0) and (0,y), and do the same for the other charge. See attached picture.

ehild
 

Attachments

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Last edited:
I'm not sure I know how to write that formula...kq1/r1 + kq2/r2 = 0?
 
First write r1. See the picture. You have a right triangle, don't you? Use Pythagorean Theorem to get r1.

ehild
 
So, kq1/(sqrt.9^2 + y^2) = 0?
 
  • #10
No, the sum of both potentials kq1/r1+kq2/r2=0!

r1= sqrt(9^2+y^2)
r2=sqrt(16^2+y^2).

ehild
 
  • #11
So, how would I solve for r1 or r2 when there are two unknowns?
 
  • #12
What are those two unknowns?

ehild
 
  • #13
y^2 and r1 for the first equation, and y^2 and r2 for the second.
 
  • #14
But you have three equations.

ehild
 
  • #15
kq1/r1 + kq2/r2 = 0 being the third?
 
  • #16
[kq1/sqrt(9^2 + y^2)] + [kq2/sqrt(16^2 + y^2)] = 0?
 
  • #17
Plug in the data for q1 and q2 and solve for y.

ehild
 
  • #18
The math isn't working for me. This is as far as I can get it:

[kq1/sqrt(9^2 + y^2)] + [kq2/sqrt(16^2 + y^2) = 0
(kq1)^2/(9^2 + y^2) = (kq2)^2/(16^2 + y^2)
[(kq1)^2*(16^2 + y^2)]/(kq1)^2 = [(kq2)^2*(9^2 + y^2)]/(kq1)^2
(16^2 + y^2)/(9^2 + y^2) = [(kq2)^2*(9^2 + y^2)]/[(kq1)^2*(9^2 + y^2)

Is this the right equation? If so, can you help me with the mathematics?
 
  • #19
Mitchtwitchita said:
The math isn't working for me. This is as far as I can get it:

[kq1/sqrt(9^2 + y^2)] + [kq2/sqrt(16^2 + y^2) = 0
(kq1)^2/(9^2 + y^2) = (kq2)^2/(16^2 + y^2)

It is right up to here.

Why don't you plug in the data for q1 and q2?

ehild
 
  • #20
Ah, I finally got it! Thank you very much for your time and patience.
 

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