Find the direction and the magnitude of the total electric field

In summary: 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
  • #1
PhilCam
46
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.
 
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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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!
 
  • #6
I have to go to bed soon, but if anyone can help me out, I'd be very grateful!
 

1. What is the concept of electric field?

The electric field is a physical quantity that describes the strength and direction of the force that a charged particle would experience at a certain point in space. It is produced by other charged particles and is responsible for the interaction between these particles.

2. How is the direction of the electric field determined?

The direction of the electric field is determined by the direction of the force that a positive test charge would experience at a given point in space. It is always directed away from positive charges and towards negative charges.

3. What is the formula for calculating the magnitude of the electric field?

The magnitude of the electric field can be calculated using the formula E = kq/r^2, where E is the electric field, k is the Coulomb's constant, q is the source charge, and r is the distance between the source charge and the point of interest.

4. How do you find the total electric field at a specific point?

To find the total electric field, you need to consider the contributions from all the individual charges in the system. You can use the principle of superposition to add up the electric fields from each charge vectorially to determine the total electric field at a specific point.

5. What are the units of electric field?

The SI unit of electric field is Newtons per Coulomb (N/C). However, it can also be expressed in volts per meter (V/m) or in terms of force per charge unit (N/C).

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