Find the magnitude of the electric field at the center

AI Thread Summary
The discussion focuses on calculating the electric field at the center of a rectangle with given charge magnitudes and dimensions. For part A, the user initially used the formula E=kq/r^2 but faced confusion regarding the contributions of different charges, ultimately arriving at a value of 143.84 N/C after corrections. In part B, the user struggled with determining the electric field components, needing guidance on using angles and the Pythagorean theorem for calculations. Another user shared a similar problem and sought clarification on the steps taken for both parts A and B. The conversation emphasizes the importance of correctly applying electric field equations and geometric principles to solve the problem.
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Homework Statement



The magnitude of each of the charges in the figure is 9.00x10^-12 C. The lengths of the sides of the rectangle are 3.00 cm and 6.00 cm. Find the magnitude of the electric field at the center of the rectangle in figures a and b.

http://www.webassign.net/CJ/18-21fig.gif



Homework Equations



E=kq/r^2



The Attempt at a Solution





For part A, I attempted to use the equation E=kq/r^2 to find the magnitude of E24 since that is all that is contributing to the electric field because of the fact that the charges at corner 1 and corner 3 are both positive and will cancel each other out. I used the pythagorean theorum to find the length of the hypotenuse (which I used as the r in the equation) of the right triangle that is created with the line between E24 and the sides of 3 cm and 6 cm to come up with a value of 6.708 which converts to .06708 meters.

E=[(8.99x10^9)(9.00x10-12)]/(.06708)^2 = 17.98 N/C I don't believe this answer is correct and this is where I have been stuck for the last few hours trying to figure out what I've done wrong.

For part B, I have no idea how to proceed because I will need the work from part A to go any further. Any guidance would be greatly appreciated!
 
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6.7 is the length across the rectangle. You want half that distance...
 
That comes out to 71.92 N/C, which still isn't the correct answer. Are my set-up and equations correct?
 
Okay, I was able to answer part a of the problem (143.84 N/C), but I'm having some trouble with the second part of the problem and hoping that someone can assist me. I believe the angle that I need to use to calculate the x and y components is 26.6 degrees. There also is not an x component to this problem, since the field only points in the y direction. But this is where I reach the point where I don't know how to proceed next?
 
Same problem, Different Values

hi,

i have a similar problem, but my problem is:
The magnitude of each of the charges in the figure is 8.00 10-12 C. The lengths of the sides of the rectangle are 3.00 cm and 4.00 cm. Find the magnitude of the electric field at the center of the rectangle in figures a and b.
I can't figure out how either parts are done. Can you show me what you did for part a and/or b?

thanks,

DeucesZ
 
Alright here is what you need to do. Once this finished you can put it in the archives for other struggling students...

In order to get the force you will need to use the electric field equation. E=kq/(distance). To get the distance you will need to do the following: convert the units to meters. Form a pythagorean triangle. Find the hypontenuse. Which for deuceszbigz is .05m Square root it. .0025. Plug into equation with charge and k. Get a value you will need to double because there is a positive charge pushing it just as much as the electric charge is pulling it.
That will get you a. For b life sucks and gets ultra tricky. You will need to do pythagorean theorem for the entire rectangle. So you have a 3,4,5 triangle and you know the hypotense. Calculate angles using SOH CAH TOA. Now you have an angle. Use your previous answer and multiply it by sin of the angle you discover (37 degrees) and multiply that by two and you got your answer.
 
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