Question about two charges on a line.

[peacemaster]

Homework Statement

A charge of 5.00mC is at the point (1.00m, 0.00m) and a charge of -4.00 is at the point (-1.00m, 0.00m). At what point, or points, on the x-axis is E=0?

Homework Equations

I decided to use the equation for voltage and set it up like this:

E = 0 = (k)(Q₁) / x + (k)(Q₂) / x


I am attempting to use the x variable as my location on the x axis.

The Attempt at a Solution

When I work the problem like I have shown above, the only possible answer to the question is infinity, and I do not believe this to be the correct answer.

Thank you guys so much for any help. I have been looking at this for a long time and can't seem to get anywhere.

Peacemaster

 

[ehild]

You have to find the point where the electric field is 0. It is not necessarily the place where the potential is 0.

ehild

 

[peacemaster]

That makes sense to me but I am not sure where to go from there. Can you suggest a formula to do that? I am still not sure what to do with the problem.

 

[ehild]

What is the electric field intensity at a distance d from a point charge of q?

ehild

 

[peacemaster]

i think you are talking about E=kQ/r^2

 

[peacemaster]

how would i set that equal to zero?

 

[ehild]

Draw a diagram. Set the positions of both charges on the x axis. Assume a test charge q=1 somewhere on the x axis, and write the forces acting on it by both charges. The sum of these forces is zero somewhere. Find that position of the test charge.

ehild

 

[peacemaster]

ehild,

I am sorry to be so stubborn, but I still don't see how to come any closer to the answer. Let's say I put a test charge of 1.00 Coulomb at the position (3.00m, 0.00m).

Now the equation I will use will be:

E = KQ₁/r² + KQ₂/r²

Going by this and using the numbers I have chosen for my test charge, I will have this:

E = (8.99E9 Nm²/C²)(-4.00E-3C)/16.00m² + (8.99E9 Nm²/C²)(5.00E-3C)/4.00m²

Sadly I do not know where to put the test charge into the equation even though I am using it's location to distinguish the radius part of the problem.

I am not usually so dense, but I am not able to bring it together this time from just the hints you have given me. Do you think you have time to walk me through the solution?

Thanks for all your help,

Peacemaster

 

[vela]

Now the equation I will use will be:

E = KQ₁/r² + KQ₂/r²

Going by this and using the numbers I have chosen for my test charge, I will have this:

E = (8.99E9 Nm²/C²)(-4.00E-3C)/16.00m² + (8.99E9 Nm²/C²)(5.00E-3C)/4.00m²

Sadly I do not know where to put the test charge into the equation even though I am using its location to distinguish the radius part of the problem.

You're looking for the location on the x-axis where E=0, right? If you placed the test charge in a different location, you'll calculate a different value of E. Somewhere on the x-axis, there's a point where E=0.

Your expression for E is not quite correct because you haven't taken into account the fact that the electric field is a vector. The electric field due to a positive charge points away from the charge; with a negative charge, the field points toward the charge. For instance, suppose you have a positive charge at x=0. At x=1, the field due to the charge would point in the +x direction. The field at x=-1 would have the same magnitude, but it would point in the -x direction.

 

[peacemaster]

thanks for your help. i asked my teacher to work it with me today and he clarified.