Where's The Electric Field 0 Problem

[japhysics]

"Where's The Electric Field 0" Problem

Homework Statement

I'm given an electron at the origin with charge "e", and an ion (with "+4 e" charge), both on the x axis. The distance between the electron and proton is given as 11 nanometers.
I need to find where the electric field will be equal to 0.

Homework Equations

E=kq/(d^2)

The Attempt at a Solution

So I know that the electric field is 0 where the individual electric fields from the electron and ion cancel each other out (right between them on the x axis).
I set negative kq/(d^2) for the electron equal to positive kq/(d^2) for the ion, so
k(4e)/(d^2) = -k(e)/[(11-d)^2] (please note the negative sign on the right hand side, because my thinking was that this is an opposite electric field, so it should have an opposite sign)
The k's and e's cancel out, but every time after doing the algebra, I simply get a quadratic equation with imaginary solutions.

Any help would really be appreciated.

 

[BruceW]

k(4e)/(d^2) = -k(e)/[(11-d)^2] (please note the negative sign on the right hand side, because my thinking was that this is an opposite electric field, so it should have an opposite sign)

this is close, but there should not be a negative sign here. When you add all the forces to get the total force, then there is a negative sign. So you had F = k(4e)/(d^2) -k(e)/[(11-d)^2] right? From here, be careful with the rearrangement.

 

[japhysics]

this is close, but there should not be a negative sign here. When you add all the forces to get the total force, then there is a negative sign. So you had F = k(4e)/(d^2) -k(e)/[(11-d)^2] right? From here, be careful with the rearrangement.

I see. So then, F=0 where the electric field is 0, so we get k(4e)/(d^2) = k(e)/[(11-d)^2]

I solved the quadratic equations and got two answers. One is two times greater than 11, the other is less than 11. I picked the one less than 11, but it still seems to be wrong (my thinking was the point where electric field is 0 has to be within the distance between the two particles).

 

[BruceW]

why does it still seem wrong? you picked the one less than 11, right? so it is between the two particles. Also, I think both the electron and the ion are negative. But your calculation takes this into account. The forces are in different directions, so the two particles must be both positive or both negative.

 

[japhysics]

why does it still seem wrong? you picked the one less than 11, right? so it is between the two particles. Also, I think both the electron and the ion are negative. But your calculation takes this into account. The forces are in different directions, so the two particles must be both positive or both negative.

I think the ion is actually positive (the problem mentioned "+4" as the charge of the ion).

I know the answer is wrong because I'm doing this on an online homework system.

Do I have to subtract what I get from 11, because of the 11-d expression?

 

[BruceW]

it will maybe help to draw a picture. put one of the charges at position zero, and label the other charge at position 11, or $L$ to make it neat. and label your 'equilibrium point' at position $x$. Then, once you decide which charge is the bigger one, it should become more clear what each of the distances are.

 

[japhysics]

One quick question; if the ion's charge is given to me as "+4e", would the charge of the electron actually be "-e" (i.e, would charge of the electron be a negative number)?

 

[Dick]

One quick question; if the ion's charge is given to me as "+4e", would the charge of the electron actually be "-e" (i.e, would charge of the electron be a negative number)?

Yes. "e" is the magnitude of the charge of the electron. It has a plus sign. Since the electron is negative, it's charge is "-e".

 

[japhysics]

So when I plug in the magnitude of e into "e" and "5e", should I take the sign into consideration?

 

[ehild]

So when I plug in the magnitude of e into "e" and "5e", should I take the sign into consideration?

You need to take the sign into consideration. And draw a picture showing the direction of the electric field due to both charges. The electric field of the positive charge is repelling, points away from the charge, the field of the negative charge is attracting.

Can the fields cancel between the charges?

ehild