Find the direction and the magnitude of the total electric field

Click For Summary

Homework Help Overview

The problem involves calculating the total electric field at the apex of an isosceles triangle formed by two point charges of equal magnitude, +q, located at the base corners. The charges are specified as 1.414 μC, and the distance from the apex to the base is given as 0.5 m, with the apex angle being 90 degrees. The gravitational forces are stated to be negligible.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field using the formula E=k(q/r^2) and the determination of the distance r based on the geometry of the triangle. There is a focus on the summation of the electric fields from both charges and the consideration of vector components, particularly the cancellation of x components and the need to account for y components.

Discussion Status

Participants are actively engaging with the problem, questioning the calculations and assumptions made regarding the electric field components. Some have requested clarification on the vector math and the geometry involved, while others are attempting to verify their calculations and reasoning.

Contextual Notes

There is a noted absence of a visual representation of the triangle, which some participants feel complicates the understanding of the problem setup. The original poster's calculations have been challenged, particularly regarding the treatment of the electric field components.

PhilCam
Messages
46
Reaction score
0

Homework Statement



1. Two point-charges of charges +q and +q are held at the corners of an isosceles rectangle
triangle, as shown in figure below. The absolute value of q is 1.414 μC. The distance d as shown in the figure is 0.5 m. The angle at A is 90. The gravitational forces are negligible.

a) Find the direction and the magnitude of the total electric field at
the apex A. (Use the coordinate system shown in figure by x and y.)

I don't have the drawing up but if you can imagine a triangle with the top angle equaling 90 degrees and the two bottom angles having charges of +q. A line descending from point A along the y-axis dissects the bottom portion of the triangle, divides it in, each side being distance d.

I know, using Summation that Etotal = E1 + E2 and that E1 and E2 are the same because they have the same charge.

I know the equation for the magnitude of an electric field is E=k(q/r^2)

To find r I used the equation .5/sin 45 = x/sin 90, ending with an r value of .707m

So I know that E = 9x10^9 ((1.414 x 10^-6)/(.707^2))

Using that equation I end with a value of 25,459V/m for E1. Using my previous equation, I find that Etotal = 25,459 + 25,459 = 50,918 v/m.

That is incorrect, can someone tell me where I slipped up? Thanks.
 
Physics news on Phys.org
PhilCam said:

Homework Statement



1. Two point-charges of charges +q and +q are held at the corners of an isosceles rectangle
triangle, as shown in figure below. The absolute value of q is 1.414 μC. The distance d as shown in the figure is 0.5 m. The angle at A is 90. The gravitational forces are negligible.

a) Find the direction and the magnitude of the total electric field at
the apex A. (Use the coordinate system shown in figure by x and y.)

I don't have the drawing up but if you can imagine a triangle with the top angle equaling 90 degrees and the two bottom angles having charges of +q. A line descending from point A along the y-axis dissects the bottom portion of the triangle, divides it in, each side being distance d.

I know, using Summation that Etotal = E1 + E2 and that E1 and E2 are the same because they have the same charge.

I know the equation for the magnitude of an electric field is E=k(q/r^2)

To find r I used the equation .5/sin 45 = x/sin 90, ending with an r value of .707m

So I know that E = 9x10^9 ((1.414 x 10^-6)/(.707^2))

Using that equation I end with a value of 25,459V/m for E1. Using my previous equation, I find that Etotal = 25,459 + 25,459 = 50,918 v/m.

That is incorrect, can someone tell me where I slipped up? Thanks.

First, "isosceles rectangle triangle" is a funny typo. Second, I really did try, but without a figure, your text is too hard for me to follow and check your math. Can you please post a figure so we can check your vector math?
 
Isosceles rectangle triangle is copied from the question! lol

I tried to print screen and hopefully this will work: http://www.freeimagehosting.net/image.php?3c6c6c04ab.jpg

Thanks for any help
 
PhilCam said:
Isosceles rectangle triangle is copied from the question! lol

I tried to print screen and hopefully this will work: http://www.freeimagehosting.net/image.php?3c6c6c04ab.jpg

Thanks for any help

That's a big help. Now show us your calc for the summation of the y components of the E field. The x components cancel obviously.
 
Oh wow, I totally ignored the Y component. I realized the X would cancel out.

Assuming that my other calculation of E = 9x10^9 ((1.414 x 10^-6)/(.707^2)) is correct, then E1 and E2 would be cos45(25,459 + 25,459) or 36004 V/m

Does that look correct?

Thanks!
 
I have to go to bed soon, but if anyone can help me out, I'd be very grateful!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
35
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 16 ·
Replies
16
Views
1K
Replies
11
Views
2K
Replies
9
Views
2K