# Finding Electric Field, Potential, and Final Velocity

1. Nov 21, 2015

### PrincessYams

1. The problem statement, all variables and given/known data

Three point charges are arranged as shown here: http://puu.sh/ltvxF.png [Broken]
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. Relevant equations

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

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

3. 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 Ex = 72000 and Ey = 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 Ey = 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.

Last edited by a moderator: May 7, 2017
2. Nov 21, 2015

### Staff: Mentor

Hello PrincessYams, Welcome to Physics Forums!

The magnitude looks okay. What is the direction?
Yes, the positive charge will create a field that is directed downwards at point A.
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.
What is the sign of the potential at A?

Last edited by a moderator: May 7, 2017
3. Nov 21, 2015

### PrincessYams

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.

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

4. Nov 21, 2015

### Staff: Mentor

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).
That looks better!

5. Nov 21, 2015

### PrincessYams

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)
E1x = 72000
E1y = 54000
E2x = -72000
E2y = 54000
E3x = 0
E3y = -56000

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

Thanks for all the help on this!

6. Nov 21, 2015

### Staff: Mentor

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:

7. Nov 21, 2015

### PrincessYams

I don't understand. Did I mix up the trig functions so that it is SUPPOSED to be:
E1 = 89900
E1x = 89900 * sin(36.9) = 54000
E1y = 89900 * cos(36.9) = 72000
E2x = -89900 * sin(36.9) = -54000
E2y = 89900 * cos(36.9) = 72000
E3x = 0
E3y = -56000

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

8. Nov 21, 2015

### Staff: Mentor

That looks better!
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.

9. Nov 21, 2015

### PrincessYams

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

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

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

10. Nov 21, 2015

### Staff: Mentor

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