Find the direction and the magnitude of the total electric field

AI Thread Summary
The discussion revolves around calculating the total electric field at the apex of an isosceles triangle formed by two point charges of equal magnitude. The user initially calculated the electric field using the formula E=k(q/r^2) and found a value of 50,918 V/m, but suspected an error. Participants pointed out the need to consider the y-components of the electric field, as the x-components would cancel each other out. The user then adjusted their calculations to account for the y-components, arriving at a new total electric field value of approximately 36,004 V/m. The conversation emphasizes the importance of vector components in electric field calculations.
PhilCam
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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.
 
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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!
 
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