Finding Electric Field, Potential, and Final Velocity

[PrincessYams]

Homework Statement

[/B]
Three point charges are arranged as shown here: http://puu.sh/ltvxF.png
a. Find the electric field at point A. (Give as vector, either as i and j or magnitude and direction.)
b. Find the potential at point A.
c. If an electron is placed at point A and released, what is the magnitude of its final velocity?

2. Homework Equations

a. E = (kq)/ r², E_y = Esin(theta), E_x = Ecos(theta)
b. V = (kq) / r
c. v = (sqrt(2P/m))

CONSTANTS:
k = 8.99 * 10⁹
m_e = 9.11 * 10^-31 kg

The Attempt at a Solution

a. For this one, I calculated all the electric fields from each point. For the left negative charge I got E = 8.9 * 10⁴, which I calculated to be E_x = 72000 and E_y = 54000. Because the right negative charge has the same charge, the electric field is the same. So, the electric fields are the same everywhere EXCEPT the x-component because that is negative, so they cancel out. Currently, I have a E_y = 72000.
For the positive charge, I got E = -5.6 * 10⁴. (I said it was negative because it is going in the negative-y direction?).
So I added all these and got 138400 upwards. My teacher said this was wrong though, where?

b. I took this piece by piece. For the left negative charge, I got -4500 ((k * -2.5 * 10⁸)/.05). For the right negative charge, I got the same thing (same charge and distance). For the middle positive charge, I got +2250 ((k * 1 * 10^-8)/ .04). Adding these together, I got 6750 V. Which, again, my teacher said was wrong. I looked it up and found no other way to do it. Even in my notes, it says to do it this way. Where did I do it wrong?

c. I did not really get where to start. I looked up in some other forums around here (will put a link when i find it) and I kept calculating it to be over the speed of light. After b is fixed, I can probably do this one. So ignore it for now.

 

[gneill]

Hello PrincessYams, Welcome to Physics Forums!

Homework Statement

[/B]
Three point charges are arranged as shown here: http://puu.sh/ltvxF.png
a. Find the electric field at point A. (Give as vector, either as i and j or magnitude and direction.)
b. Find the potential at point A.
c. If an electron is placed at point A and released, what is the magnitude of its final velocity?

2. Homework Equations

a. E = (kq)/ r², E_y = Esin(theta), E_x = Ecos(theta)
b. V = (kq) / r
c. v = (sqrt(2P/m))

CONSTANTS:
k = 8.99 * 10⁹
m_e = 9.11 * 10^-31 kg

The Attempt at a Solution

a. For this one, I calculated all the electric fields from each point. For the left negative charge I got E = 8.9 * 10⁴, which I calculated to be E_x = 72000 and E_y = 54000. Because the right negative charge has the same charge, the electric field is the same. So, the electric fields are the same everywhere EXCEPT the x-component because that is negative, so they cancel out. Currently, I have a E_y = 72000.

The magnitude looks okay. What is the direction?

For the positive charge, I got E = -5.6 * 10⁴. (I said it was negative because it is going in the negative-y direction?).

Yes, the positive charge will create a field that is directed downwards at point A.

So I added all these and got 138400 upwards. My teacher said this was wrong though, where?

Check that you're summing the values taking into account their directions. It's a good idea to sketch in the field vectors from the individual charges on your diagram to guide and confirm your math.

b. I took this piece by piece. For the left negative charge, I got -4500 ((k * -2.5 * 10⁸)/.05). For the right negative charge, I got the same thing (same charge and distance). For the middle positive charge, I got +2250 ((k * 1 * 10^-8)/ .04). Adding these together, I got 6750 V. Which, again, my teacher said was wrong. I looked it up and found no other way to do it. Even in my notes, it says to do it this way. Where did I do it wrong?

What is the sign of the potential at A?

 

[PrincessYams]

Check that you're summing the values taking into account their directions. It's a good idea to sketch in the field vectors from the individual charges on your diagram to guide and confirm your math.

Oh! That is what I did. Sketching the diagram really did help. So now the Ex is gone. It was the 5400 + 5400 + (-5600) I was supposed to do. So I got 52000 upwards/positive y-direction? Thanks! I should mark these more clearly in my work and check over small mistakes like that.

What is the sign of the potential at A?

I forgot Potential can be negative! Thanks. So would it just be V = - 6750 V?

 

[gneill]

Oh! That is what I did. Sketching the diagram really did help. So now the Ex is gone. It was the 5400 + 5400 + (-5600) I was supposed to do. So I got 52000 upwards/positive y-direction? Thanks! I should mark these more clearly in my work and check over small mistakes like that.

Those "5400" values look like suspiciously x-component values for the negative charges, but a power of 10 too small (values had a 10⁴ associated with them when you showed your work). Check that you're using the right trig function to determine your components. (Hint: You can use the legs of the triangles directly to calculate the trig ratios).

I forgot Potential can be negative! Thanks. So would it just be V = - 6750 V?

That looks better!

 

[PrincessYams]

Those "5400" values look like suspiciously x-component values for the negative charges, but a power of 10 too small (values had a 104 associated with them when you showed your work). Check that you're using the right trig function to determine your components. (Hint: You can use the legs of the triangles directly to calculate the trig ratios).

Oh, yeah oops. That was a typo. To clarify what I got (for my sake):
E₁ = 89900 (to get the x- and y-components I multiplied this by cos(36.9) and sin(36.9) respectively)
E_1x = 72000
E_1y = 54000
E_2x = -72000
E_2y = 54000
E_3x = 0
E_3y = -56000

All x-values = 0
All y-values = 52000

Thanks for all the help on this!

 

[gneill]

I think you've reversed the trig functions. That angle, 36.9°, would be an angle at the bottom of a 3-4-5 triangle in your figure. For the y-component of $E_1$ or $E_2$ you're looking for their projection onto the vertical leg of the triangle. That would make the desired trig function a cosine:

 

[PrincessYams]

I don't understand. Did I mix up the trig functions so that it is SUPPOSED to be:
E₁ = 89900
E_1x = 89900 * sin(36.9) = 54000
E_1y = 89900 * cos(36.9) = 72000
E_2x = -89900 * sin(36.9) = -54000
E_2y = 89900 * cos(36.9) = 72000
E_3x = 0
E_3y = -56000

Or am I using the wrong angle entirely?[/SUB]

 

[gneill]

I don't understand. Did I mix up the trig functions so that it is SUPPOSED to be:
E₁ = 89900
E_1x = 89900 * sin(36.9) = 54000
E_1y = 89900 * cos(36.9) = 72000
E_2x = -89900 * sin(36.9) = -54000
E_2y = 89900 * cos(36.9) = 72000
E_3x = 0
E_3y = -56000

That looks better!

Or am I using the wrong angle entirely?[/SUB]

You can use either angle in the triangle, but you need to use the appropriate trig function to pull out the required component.

For example, in the diagram that I posted you can see that the projection of the $E_1$ vector onto the vertical 4 cm leg of the triangle requires the cosine of the angle θ. If you had chosen to use the other angle (which is about 53°) then you would need to use the sine to find that vertical component.

 

[PrincessYams]

For example, in the diagram that I posted you can see that the projection of the E1E_1 vector onto the vertical 4 cm leg of the triangle requires the cosine of the angle θ. If you had chosen to use the other angle (which is about 53°) then you would need to use the sine to find that vertical component.

Oh, I see. The example I was looking at must have done something like that and I didn't notice. I see now.

That looks better!

Thanks for all the help!
72000 + 72000 - 56000 = 88000

Finding the final velocity should just be:
v = √ (2Q(change in V)) / m)
?

 

[gneill]

72000 + 72000 - 56000 = 88000

Sure. Be sure to include units on results! Electric field is usually specified in N/C or V/m.

Finding the final velocity should just be:
v = √ (2Q(change in V)) / m)
?

Yup.