Finding the Magnitudes of Charges in a Rectangle Using Electric Field Analysis

In summary, the book is terrible at explaining this stuff, so I was hoping someone here could help me out. I tried to solve for q1 and q2 using vector calculus, but I ran into trouble when I tried to find the resultant of the vectors. I think it would be helpful to show your work if you decide to go that route.
  • #1
ToyMachine
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These have been driving me crazy. The book is terrible at explaining this stuff, so I was hoping someone here could help me out.

Homework Statement


A rectangle has a length of 2d and a height of d. Each of the following three charges is located at a corner of the rectangle: +q1 (upper left corner), +q2 (lower right corner), and -q (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of q1 and q2. Express your answers in terms of q.

Homework Equations


Coloumb's Law: k * (|q1| * |q2| / r2)
Where k= a proportionality constant ~ 8.99 * 109 N * m2/C2

Electric Field Definition: E = F / q0
Where E is the net electric field at a point, and F is the force experienced by a small test charge represented by q0

The Attempt at a Solution


I really got nowhere trying to find this solution, but here's what I tried:
--> I defined the upper right corner where the net electric field is zero as point T (for easy reference).
--> I represented the forces exerted on point T as three vectors, all with unknown magnitudes. The vector created by q1 had a direction of 0º; the vector created by q2 had a direction of arctan(2)~63.43º (reasoning below), and the vector created by -q had a direction of 270º.
--> I obtained the direction of the vector created by q2 by drawing a right triangle with leg lengths 1 and 2 and solving for the angle opposite the side with length 2. The leg lengths were obtained from the given data that the sides of the rectangle are d and 2d.
--> At this point, I realized I was completely on the wrong track. I was planning on solving for the resultant of these three vectors, but I realized that it was already given in the problem that the resultant is, in effect, zero. Thus, in my line of thought, the resultant of the vectors produced by q1 and q2 (the positive charges) must be equal in magnitude and opposite in direction of the vector produced by -q. However, because the two vectors produced by these positive charges are at 0º and ~63.43º, they cannot produce a resultant at 90º, which would be needed in order to have the opposite direction of the vector created by then negative charge. Thus, to me, the problem appears impossible, unless q1 or q2 were allowed to be negative, which I don't believe they are.


Thanks a ton for any help you can provide!
 
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  • #2
I think the way to go with this is to think in terms of vector components. Since the charge in the upper left only contributes along the x-axis and the one in the lower right contributes along the y-axis then the E-field from the other charge must cancel both these components for the E-field to be zero.
 
  • #3
Alright, well it turns out I wasn't the only one with a problem on this one.
We went over this problem at the very end of the class period today, and the bell rang before we had a chance to finish it, and given the nature of the class and teacher I doubt we'll be coming back to it (much to my disappointment).

I did as you suggested and took the vector components. However, given the data, I was only able to obtain a ratio of the forces exerted on T.
I came up with something to the effect of: for every 1 unit of force exerted by -q on a test charge at point T (again, that's the upper-right hand corner, where the net electric field is zero), charge q1 exerts about .8 units of force, and charge q2 exerts about .4 units of force. (I'm not looking at the problem currently, so I may have mixed up q1 and q2, but I'm pretty sure the ratio is close to correct (sorry I couldn't remember more numbers after the decimal))

Now, the trouble is going from this conclusion to the data that the question asks for. Given that these are the ratios of the forces exerted and knowing the relative distances between all four of the points, what are the magnitudes of q1 and q2 in terms of q (and, I assume, d)?
 
  • #4
If you show your work it'll be easier to see where you're making a mistake. Although if you do follow my advice it should be pretty easy to come up with an answer.
 

1. What is an electric field?

An electric field is a physical field that surrounds electrically charged particles and exerts a force on other charged particles within its range. It is represented by a vector quantity and is measured in units of volts per meter (V/m).

2. How do you calculate the strength of an electric field?

The strength of an electric field can be calculated using the formula E = F/q, where E is the electric field strength, F is the force exerted on the charge q, and q is the magnitude of the charge. Alternatively, it can also be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the source charge, and r is the distance from the source charge.

3. What is the difference between an electric field and an electric potential?

An electric field represents the force per unit charge exerted by a source charge, while an electric potential represents the potential energy per unit charge of a charged particle in an electric field. In other words, an electric field is a vector quantity while an electric potential is a scalar quantity.

4. How do electric fields interact with conductors and insulators?

Electric fields can penetrate through insulators, but they cannot penetrate through conductors. In conductors, electric charges are free to move and will redistribute themselves in such a way that the electric field inside the conductor is zero. In insulators, electric charges are bound and cannot move freely, so the electric field will remain unchanged.

5. What are some real-world applications of electric field problems?

Electric field problems are important in understanding and designing electrical circuits, as well as in the operation of electronic devices such as radios, computers, and cell phones. They are also crucial in fields such as electromagnetism, telecommunications, and particle physics.

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