# How to find the magnitude of a point charge given 2 others?

• shqiptargirl
In summary, the charges on the three objects are as follows: q1 is 2.00 µC and its sign is unknown; q2 is not known, but it must be positive because the y component of the net force vector of q3 is zero; q3 is +4.00 µC and the net force on it is entirely in the negative x-direction.
shqiptargirl

## Homework Statement

Three charges are placed as shown in the figure below. The magnitude of q1 is 2.00 µC, but its sign is not known. The charge and sign of q2 is not known. We do know that q3 is +4.00 µC and the net force on q3 is entirely in the negative x-direction.

a.) Deduce the signs of charges q1 and q2.
b.) Figure out the magnitude of q2

## Homework Equations

Coulomb's law:

F = (K)*(q1)(q2)/d^2

## The Attempt at a Solution

I know how to determine the charges of q1 and q2. Since the y component of the net force vector of q3 equals 0, then q1 must be negative and q2 must be positive.

Besides that, I have no idea on how to start figuring out the magnitude of q2

shqiptargirl said:
Since the y component of the net force vector of q3 equals 0, then q1 must be negative and q2 must be positive.
Well, it could have been the other way around, but then the net force would have been in the +ve x direction.
Your observation about net zero in the y direction tells you more than you've used. What is the y direction force from q1 on q3?

haruspex said:
Well, it could have been the other way around, but then the net force would have been in the +ve x direction.
Your observation about net zero in the y direction tells you more than you've used. What is the y direction force from q1 on q3?

The y direction force from q1 on q3 is 23.15 N. So The y direction force from q2 from q3 must be -23.15 N ? The problem is I don't know how to use this information to solve for the magnitude of q2.

Also, what does this mean, "+ve x direction" ? I didn't understand your comment on how it could've been the other way around (sorry) :(

shqiptargirl said:
The y direction force from q1 on q3 is 23.15 N
I don't know k off the top of my head so I can't quickly check that. For the purposes of the question, the value of k is immaterial, so we can pretend it's 1 whatever, consistent with expressing distances in cm and charges in µC.
shqiptargirl said:
The y direction force from q2 from q3 must be -23.15 N ?
Yes, that would follow. You know the magnitude of q3, and the position of q2 in relation to it, so you can write an equation for that force as a function of q2's magnitude.
shqiptargirl said:
what does this mean, "+ve x direction"
Sorry, +ve means positive, -ve means negative. I thought these were very standard.
shqiptargirl said:
I didn't understand your comment on how it could've been the other way around
You could set q1 positive and q2 negative and still arrange that there's not net Y field at q3. So you need to use the fact that the X field at q3 is to the left to determine which is positive and which is negative.

You said this... "Yes, that would follow. You know the magnitude of q3, and the position of q2 in relation to it, so you can write an equation for that force as a function of q2's magnitude."

I don't understand this part. I am lost on how to do this.
And k btw is about 9 *10 ^9, atleast that is what I have been using.

shqiptargirl said:
Hm, but if you flipped them around, wouldn't the arrow be pointing in the opposite direction? If you look at the diagram the arrow is pointing to the left. If you made q1 positive and q2 negative, the arrow would be pointing to towards the right, no?
Yes, that's exactly what I'm saying. In your original post you wrote
shqiptargirl said:
Since the y component of the net force vector of q3 equals 0, then q1 must be negative and q2 must be positive.
as though the zero y component was enough to determine that q1 must be negative and q2 must be positive. That observation only tells you they are of opposite sign. To deduce that it's q1 that's negative you also need to use the direction in which the arrow points.
You probably worked all that through when you made the post, but misrepresented your own logic.

It really is a very small point, and I'd rather get on with finding q2.

I know, I am sorry. I didn't see the last part of your reply because of the computer screen I am using.

## 1. How do I find the magnitude of a point charge if I know the charges of two other points?

To find the magnitude of a point charge, you will need to use Coulomb's Law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. You can use this equation to solve for the magnitude of the point charge by plugging in the values of the other two charges and the distance between them.

## 2. What units are used to measure the magnitude of a point charge?

The magnitude of a point charge is typically measured in Coulombs (C). However, depending on the problem, you may need to convert the charge to its SI unit, the fundamental unit of charge, which is the Coulomb.

## 3. Can I use Coulomb's Law to find the magnitude of a point charge in a three-dimensional space?

Yes, Coulomb's Law can be used to calculate the magnitude of a point charge in three-dimensional space. You will need to use the distance formula to find the distance between the point charge and the other two charges in order to plug it into the equation.

## 4. What if the point charge is located at the midpoint between the two other charges?

If the point charge is located at the midpoint between the other two charges, then the distance between them will be half of the total distance. You can plug this value into the equation to solve for the magnitude of the point charge.

## 5. Are there any other factors that can affect the magnitude of a point charge?

Other factors that can affect the magnitude of a point charge include the medium surrounding the charges, as well as any other external forces acting on the charges. These factors may alter the distance or the charges themselves, and therefore affect the overall magnitude of the point charge.

• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
261
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
17
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
21
Views
817