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

In summary, the homework statement states that 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? The Attempt at a Solution states that 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
  • #1
Mitchtwitchita
190
0

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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  • #2
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
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
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
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
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.
 

1. What is the x-axis in relation to electric potential?

The x-axis in electric potential is a representation of distance or position in space. It is typically used to measure the distance from a reference point to a specific location where the electric potential is being calculated.

2. How is the electric potential on the x-axis calculated?

The electric potential on the x-axis is calculated using the formula V = kQ/x, where V represents the electric potential, k is the Coulomb's constant, Q is the electric charge, and x is the distance from the reference point.

3. What units are used to measure electric potential on the x-axis?

The units used to measure electric potential on the x-axis are volts (V) or joules per coulomb (J/C).

4. Can the electric potential on the x-axis be negative?

Yes, the electric potential on the x-axis can be negative. This occurs when the reference point has a higher electric potential than the point being measured. It indicates that work must be done to move a positive charge from the reference point to the point being measured.

5. How does the electric potential on the x-axis change with distance?

The electric potential on the x-axis follows an inverse relationship with distance. As the distance increases, the electric potential decreases, and vice versa. This is due to the fact that the electric field strength weakens with distance and therefore the work required to move a charge also decreases.

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