# Very hard Coulomb Force with two charges

• Ben000
In summary, the conversation is about calculating the electric field and Coulomb force experienced by two charges (q1 = -15uC and q2 = 3uC) separated by a distance of 3m, and finding the location x where the electric field is zero. The equations used are Coulomb's Law and potential for a point charge. To solve for x, the tutor suggests setting the equations for the Coulomb force from each charge equal to each other and solving for x. After using the quadratic formula, the possible values for x are -2.4 and 0.9. However, the student is still confused and unsure of which value to use and how to check if it is correct.
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.

## Homework Statement

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.

## Homework 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.

## 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.

Ben000 said:
Coulomb's Law for the Electric field:

$$\vec{F}=k\frac{q_1q_2}{r^2}\hat{r}$$ or $$\vec{F} = \frac{q_1q_2}{4\pi\epsilon_0r^2}\hat{r}$$

and potential for a point charge:

$$V = k\frac{q}{r}$$ or $$V = \frac{q}{4\pi\epsilon_0r}$$

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:

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

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

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

AM

Ben000 said:
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.

## Homework Statement

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.

## Homework Equations

Coulomb's Law:
F=k q1q2/r2 or 1/(4πϵ0) q1q2/r^2
Electric field and potential for a point charge:
E=k q/r2 or 1/(4πϵ0) q/r2k
V=q/r or 1/(4πϵ0) q/r

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

## 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*10^9(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*10^9(|15uC*3uC|)/3^2
F=5^-12    How did you get this answer? 1μC = 1*10-6C .

Furthermore, the problem does not ask for the force that the charges exert on each other.

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.
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:
Andrew Mason;4246133 Set [itex said:
\vec{E} = 0[/itex] and solve for x. (Hint: You will have to solve a quadratic equation).

AM

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?

Ben000 said:
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.
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.

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

(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

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.

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?
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:

## 1. What is the Coulomb force?

The Coulomb force, also known as the electrostatic force, is a fundamental force of nature that describes the interaction between electrically charged particles. It is responsible for the attraction and repulsion between two charged objects.

## 2. How is the Coulomb force calculated?

The Coulomb force is calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is F = k * (q1 * q2) / r^2, where F is the force, k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

## 3. What is the unit of measurement for the Coulomb force?

The unit of measurement for the Coulomb force is newtons (N). This is the same unit used to measure other types of forces, such as gravity.

## 4. How does the Coulomb force change with distance?

The Coulomb force follows an inverse square law, which means that as the distance between two charged particles increases, the force between them decreases. This means that the force is strongest when the particles are close together and gets weaker as they move farther apart.

## 5. Can the Coulomb force be attractive and repulsive at the same time?

No, the Coulomb force can only be either attractive or repulsive between two charged particles. If the charges are of opposite signs, the force will be attractive, and if they are of the same sign, the force will be repulsive. This is based on the principle that like charges repel and opposite charges attract.

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