Electric Field Question: Finding Zero Point Between Two Charges on Y-Axis

In summary, the conversation is about finding the point at which the total electric field is zero between two point charges along the y-axis. The equation set up to solve for this point is q1/(r-6)^2 = q2/(r+4)^2, but there is a typo in the equation. After correcting the typo, the equation becomes sqrt(q1/q2) = (6-r)/(r+4). To solve for r, one must multiply both sides by (r+4), resulting in a simple linear equation.
  • #1
thebigbluedeamon
7
0
I need a little guidance in this problem...

Two point charges lie along the y-axis. A charge of q1=-9 mu*C is at y=6.0m, and a charge of q2=-8.0 mu*C is at y=-4.0 m. Locate the point (other than infinity) at which the total electric field is zero.

So,

I made the statement

E1y = -E2y

and

Ke (q1/(r-4)^2) = Ke (q2/(r+4)^2)
or
q1/(r-6)^2 = q2/(r+4)^2

But that makes it very hard to solve for r. Is that equation set up correctly? If so, what is the easiest way, algebraicly, to solve for r.
 
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  • #2
thebigbluedeamon said:
Ke (q1/(r-4)^2) = Ke (q2/(r+4)^2)
or
q1/(r-6)^2 = q2/(r+4)^2

But that makes it very hard to solve for r. Is that equation set up correctly? If so, what is the easiest way, algebraicly, to solve for r.
I think you have a typo in the equation above. Take the square root of both sides and simplify.
 
  • #3
e(ho0n3 said:
I think you have a typo in the equation above. Take the square root of both sides and simplify.

I did have a typo...It was supposed to be a 6 instead of a 4 in the first equation.

Let me try this and hopefully I can solve.
 
  • #4
Okay...so I took the square root of both sides of the equation, but that doesn't seem to get me very far.

I end up with:

sqrt(q1/q2) = (r-6)/(r-4)

And frankly I don't know how to solve for r in this situation. I could use Maple or something, but it seems that this problem shouldn't require that. I think I might have got the initial set up wrong.
 
  • #5
thebigbluedeamon said:
sqrt(q1/q2) = (r-6)/(r-4)
Two problems: (1) another typo, and (2) when you take square roots you better be sure your answer is positive.

Your equation should be: sqrt(q1/q2) = (6-r)/(r+4)
And frankly I don't know how to solve for r in this situation.
Start by multiplying both sides by (r+4). It's a simple linear equation.
 
  • #6
Of course it is. I just had a mental block. I wouldn't have caught the "(6-r)" though. Thanks for your help.
 

1. What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle or object. It is responsible for the electric force that acts on other charged particles within the field.

2. How is an electric field created?

An electric field is created by placing a charged particle or object in a space. The magnitude and direction of the electric field is determined by the charge of the particle or object.

3. How is the strength of an electric field measured?

The strength of an electric field is measured in units of volts per meter (V/m). This unit describes the amount of force per unit charge that a charged particle would experience within the electric field.

4. How does distance affect the strength of an electric field?

The strength of an electric field is inversely proportional to the square of the distance between the charged particle and the object creating the field. This means that as the distance between the two increases, the strength of the field decreases.

5. What are some real-world applications of electric fields?

Electric fields have many practical applications, such as in electronic devices, power generation and transmission, and medical technologies such as MRI machines. They also play a crucial role in understanding and manipulating particles at a microscopic level in fields such as physics and chemistry.

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