# Homework Help: Very hard Coulomb Force with two charges

1. Jan 27, 2013

### Ben000

I'm new to forums and after I typed all this out and tried to submit, it cleared everything. I really need to get this problem finished tonight.

1. The problem statement, all variables and given/known data
You have two charges (q1 = -15uC and q2 = 3uC) separated by a distance (d = 3m). We want to calculate the electric field, E (vector quantity), at a location x relative to charge q2 located on a line connecting the two charges. Note that x could be anywhere on that line. Calculate the Coulomb force experienced by each charge. Give both magnitude and direction.

Also, Find the location x, where E = 0.

2. Relevant equations
Coulomb's Law:
F=k q1q2r2 or 14πϵ0 q1q2r2
Electric field and potential for a point charge:
EV==k qr2 or 14πϵ0 qr2k qr or 14πϵ0 qr

I don't know how to enter them in a fancy way, I tried copy pasting them from the sticky.

3. The attempt at a solution

So I know I have to first come up with x before I can solve for E. But if I set up the equation it seems impossible.

0=8.99^9(3uC/x)

There's no way to get x on the other side without it equaling zero!

So I don't know how to move forward in the problem.

For the charge part, I don't really understand magnitude and direction.

As a guess I would say that both charges experience an equal force from each other, only differing in direction. Because the negative one would go towards the positive.

F=8.99^9(|15uc*3uc|)/3^2
F=5^-12

That sounds like a really small number and I don't know what unit it would be.

PLEASE help me I don't want to fail my physics. I like this stuff in concepts but I can't do the problems.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 27, 2013

### Andrew Mason

Put the origin at q2 and use the Coulomb's law equation (I have fixed it up for you) to calculate the field (force per unit charge) from each charge at a point x (units are MKS):

$\vec{E_2}=k\frac{q_2}{(distance from q_2)^2}\hat{i}$

$\vec{E_1}=k\frac{q_1}{(distance from q_1)^2}\hat{i}$

The field at x is just:

$\vec{E} = \vec{E_1} + \vec{E_2}$

Set $\vec{E} = 0$ and solve for x. (Hint: You will have to solve a quadratic equation).

AM

3. Jan 27, 2013

### SammyS

Staff Emeritus
Hello Ben000. Welcome to PF !

(There are some corrections & imbedded comments above, in RED)

The problem, as stated, doesn't say whether q1 (the -15μC charge) is at x = 3m or at x = -3 m .

The electric field points away from a positive charge and points toward a negative charge.

Assuming that q1 is at x = 3m:
For x > 3, Etotal = E2 - E1 .

For 0 < x < 3, Etotal = E2 + E1 .

For x < 0, Etotal = E1 - E2 .

That's just a bare start to solving this problem.

Last edited: Jan 28, 2013
4. Jan 27, 2013

### Ben000

I just talked with my tutor before he had to go to bed but here is what he told me to do:

"Calculate the coulomb force from either charge in terms of x. For q2 you can let r=x and for q1 you'll want r= 3m - x. Set the two equations equal to each other and solve for x. "

(k*q2)/x^2 = -(k*q1)/(3-x)^2

"You need the sum of E to be zero. so the E's need to be equal and opposite.
The point where the magnitudes of the electric fields is the same, and the directions are opposite will be zero."

(k*q1)/(3-x)^2 = (k*q2)/x^2

q1/(3-x)^2 = q2/x^2

q1*(x^2) = q2*(3-x)^2

-15/(3-x)^2 = -3/x^2 =>> 5/(3-x)^2 = 1/x^2

5*x^2 = (3-x)^2

0 = 9-6x-4x^2

x= -2.4 & 0.9

x is a distance from q2
the negative number will be to the right hand side, and the positive to the left of q2
-----------q1(-15uC)--------x=0.9-----q2(3uC)-------x=-2.4

Still after all of this I am very confused and feel like I cannot do this problem.

I don't know how to check if my x is correct or which one to use.

AM,
What does it mean distance from q1/q2 in the equations you gave? I thought it would be 0 in both cases because the points are on themselves, so d = 0?

5. Jan 28, 2013

### Andrew Mason

I am not sure how you are setting this up but if q2 is at the origin and q1 is at +3m, the displacement from q1 to x is x - (+3). If q1 is at -3m then the displacement from x to q1 is x - (-3) = x + 3.

Correct: $\vec{E} = \vec{E_1} + \vec{E_2} = 0$

One has to be careful with the direction of the field as well as the correct expression for the magnitude of the field. The direction of the field of a positive charge is radially away from the charge. If you are putting q1 at x = 3m and q2 at x=0, then:

for x>0, (x-3) is the displacement relative to q1 and x is the displacement relative to q2 and the field is:

$\vec{E} = \frac{k*q1}{(x-3)^2}\hat{i}+ \frac{k*q2}{x^2}\hat{i}$ for x>3 and

$\vec{E} = \frac{k*q1}{(x-3)^2}\hat{-i}+ \frac{k*q2}{x^2}\hat{i}$ for 0<x<3

for x<0, $(|x|+3)\hat{-i}$ is the displacement relative to q1 and $|x|\hat{-i}$ is the displacement relative to q2 and the field is:

$\vec{E} = \frac{k*q1}{(|x|+3)^2}\hat{-i}+ \frac{k*q2}{|x|^2}\hat{-i}$

So you have to solve three different equations here: one for x>3, one for 0<x<3 and one for x<0.

For x>0, the roots for |x| are negative (which means there is no solution):

So try for x<0. This should give you two positive roots for |x|. Since x<0 you have to multiply by -1 to get the solutions.

The force would be undefined for a position d=0, assuming point charges. x just represents an arbitrary position on the x axis. If q1 is at x=0 and q2 is at x=3 then E is undefined at x=0 and x=3.

AM

Last edited: Jan 28, 2013