Electric Potential x-axis

1. Jun 22, 2010

Mitchtwitchita

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

V = kq/r

3. 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?

2. Jun 22, 2010

ehild

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. Jun 22, 2010

Mitchtwitchita

So, I can take any arbitrary point on the y-axis, and it will eventually lead me to the right answer?

4. Jun 22, 2010

ehild

Yes, if your procedure is correct. Let me see how you proceed.

ehild

5. Jun 22, 2010

Mitchtwitchita

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. Jun 22, 2010

ehild

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

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Last edited: Jun 22, 2010
7. Jun 22, 2010

Mitchtwitchita

I'm not sure I know how to write that formula...kq1/r1 + kq2/r2 = 0?

8. Jun 22, 2010

ehild

First write r1. See the picture. You have a right triangle, don't you? Use Pythagorean Theorem to get r1.

ehild

9. Jun 22, 2010

Mitchtwitchita

So, kq1/(sqrt.9^2 + y^2) = 0?

10. Jun 22, 2010

ehild

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. Jun 22, 2010

Mitchtwitchita

So, how would I solve for r1 or r2 when there are two unknowns?

12. Jun 22, 2010

ehild

What are those two unknowns?

ehild

13. Jun 22, 2010

Mitchtwitchita

y^2 and r1 for the first equation, and y^2 and r2 for the second.

14. Jun 22, 2010

ehild

But you have three equations.

ehild

15. Jun 22, 2010

Mitchtwitchita

kq1/r1 + kq2/r2 = 0 being the third?

16. Jun 22, 2010

Mitchtwitchita

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

17. Jun 22, 2010

ehild

Plug in the data for q1 and q2 and solve for y.

ehild

18. Jun 22, 2010

Mitchtwitchita

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. Jun 22, 2010

ehild

It is right up to here.

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

ehild

20. Jun 22, 2010

Mitchtwitchita

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