Electric field and rectabgle

In summary, the conversation discusses the arrangement of three positive charges in a rectangle and finding the magnitude of the electric field at the fourth corner of the rectangle. The solution involves using the equation E = kq / r^2 and breaking down the field components from each charge. After realizing some false assumptions, the correct solution is found using the ratios of the sides of the rectangle.
  • #1
itryphysics
114
0

Homework Statement


Three positive charges are arranged in a rectangle. The charge in the bottom left corner is +5.63 nC, in the top left corner is +5.64 nC, and in the bottom right corner it is +11.28 nC. The sides have a length of 0.639 m and .213 m. Find the magnitude of the electric field at the fourth corner of the rectangle. Answer in units of N/C.


Homework Equations


E = kq / r^2


The Attempt at a Solution


E for bottom left : k(q) / (.67356m)^2 = 111.53 N/C

E for top left : kq/ (.213m)^2 = 1117.28 N/C

E for bottom right : kq / (.639m)^2 = 248.284 N/C

Then I drew these Electric fields on an x-y axis and they all ended up in quadrant one. So I then found the x an dy components of each. Ex total was 1196.14 and Ey total was 327.148
Finally I found the Electric field for 4th corner by using the x and y compoenents and it turned out to be 1240.07 N/C .
for some reason I am not wholly assured that my answer is correct. Can you please look over and tell me if there is anything wrong with my problem solving approach. Thank you very much =]
 
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  • #2
sure enough I entered the answer and it was wrong =( . I am now in dire need of help
 
  • #3
itryphysics said:

The Attempt at a Solution


E for bottom left : k(q) / (.67356m)^2 = 111.53 N/C

E for top left : kq/ (.213m)^2 = 1117.28 N/C

E for bottom right : kq / (.639m)^2 = 248.284 N/C

Good :smile:
So I then found the x an dy components of each. Ex total was 1196.14 and Ey total was 327.148

That's not what I get. Perhaps you should show me what you are getting for the x and y components of the field from the bottom left charge.
 
  • #4
Thank you for cheking my work . My approach was correct. The way I had drawn my diagram of the charges was incorrect, which led me to make some false assumptions. For example I was using 45 degrees as my angle when i was finding components. However that is incorrect because the figure is a rectabgle not a square so instead I had to use the ratios of the sides in order to calculate my components.
Thank you once again!
 
  • #5
Welcome :smile:
 

Related to Electric field and rectabgle

What is an electric field?

An electric field is a region around a charged particle or object where other charged particles will experience a force. It is created by the presence of a charged particle and can be either positive or negative.

How is the strength of an electric field measured?

The strength of an electric field is measured by its electric field intensity, which is the force per unit charge at a given point in the field. It is represented by the symbol E and is measured in newtons per coulomb (N/C).

What is a rectangle in relation to electric fields?

A rectangle is a shape that can be used to represent the cross-sectional area of an electric field. It is often used in diagrams and calculations to demonstrate the direction and magnitude of the electric field at different points.

How does the size and shape of a rectangle affect the electric field?

The size and shape of a rectangle can affect the electric field by changing the electric field lines and the distribution of the electric field intensity. For example, a larger rectangle will have a larger cross-sectional area for the electric field to act on, resulting in a stronger electric field.

How is the electric field inside a rectangle distributed?

The electric field inside a rectangle is distributed evenly across the cross-sectional area. This means that the electric field intensity is the same at all points within the rectangle, unless there are other charged particles or objects present that may affect the field.

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