Finding the field strength of an EF

[Watsonb2]

Homework Statement

A square, with charges at each corner. The upper left (1) has a charge of +q. The upper right (2) has a charge of -2q. The bottom right (3) has a charge of +2q. The bottom left (4) has a charge of -q. The square has side length a.

a = 5.20 cm and q = 11.8 nC

Find the EF at the center of the square.

Homework Equations

This was the equation that my prof gave us to deal with an observation point (in this case the center of the square) introduced into an Electric Field:

[tex]\vec{E}[/tex]([tex]\vec{r}[/tex] = [k (Q) ([tex]\vec{r}[/tex] - [tex]\vec{R}[/tex]] / [([tex]\vec{r}[/tex] - [tex]\vec{R}[/tex])^3]

Where r = the actual distance from the source charge to the observation point and R = the distance of the source charge from the origin.

The Attempt at a Solution

I started by determining the distances between the source charges and the observation point, which all were a/[tex]\sqrt{2}[/tex], by pythagorean theorem.

I plugged this into the equation and got an answer of zero, which immediately flagged my attention, but I can't see why it came out that way. I then plugged the numbers into the equation for the second charge and got an answer of -1.57 x 10^14. This seemed very large, even though the charge of q is fairly large to begin with...

I left off there, maybe you guys could help me out and point out where I'm making my error...

Thanks,

-B

 

[rl.bhat]

Electric fields due to charges 1 and 3 are in the opposite direction. The net field is towards charge 1. Similarly net field due charges due to 2 and 4 is towards charge 4. These two net fields are perpendicular to each other. Now find the resultant field due to all the four charges at the center.

 

[nlightner]

I would first like to commend the student for attempting to solve the problem and for reaching out for help when they encountered difficulties. It is important to seek assistance and clarification when faced with challenges in scientific problem-solving.

In this case, it seems that the student may have made a mistake in their calculations. The equation provided by the professor is correct, but it is important to plug in the correct values for each variable. In this case, the distance between the source charges and the observation point should be a/√2, as the student correctly noted. However, the value for R should be the distance from the origin to the source charge, which is simply a. This is because the equation is using vector notation, so the vector from the origin to the source charge is represented as (a, 0) and the vector from the origin to the observation point is represented as (a/√2, a/√2).

Using these values, the equation should give an electric field of -1.51 x 10^14 N/C at the center of the square. This may still seem like a large number, but it is important to remember that electric field strength is dependent on the distance from the source charge, as well as the magnitude of the charge. In this case, the charges are relatively close to the observation point, so the electric field strength will be relatively high.

In conclusion, it is important to double check calculations and make sure all variables are plugged in correctly when solving problems in physics. It is also helpful to have a good understanding of the concepts and equations being used, which can be achieved through practice and seeking clarification when needed.