Electric Potential x-axis

  • #1

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?
 

Answers and Replies

  • #2
ehild
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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
 
  • #3
So, I can take any arbitrary point on the y-axis, and it will eventually lead me to the right answer?
 
  • #4
ehild
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Yes, if your procedure is correct. Let me see how you proceed.

ehild
 
  • #5
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.
 
  • #6
ehild
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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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  • #7
I'm not sure I know how to write that formula...kq1/r1 + kq2/r2 = 0?
 
  • #8
ehild
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First write r1. See the picture. You have a right triangle, don't you? Use Pythagorean Theorem to get r1.

ehild
 
  • #9
So, kq1/(sqrt.9^2 + y^2) = 0?
 
  • #10
ehild
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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
ehild
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What are those two unknowns?

ehild
 
  • #13
y^2 and r1 for the first equation, and y^2 and r2 for the second.
 
  • #14
ehild
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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
ehild
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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
ehild
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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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