# The Electric Field problem

1. Feb 11, 2007

### FSNxPhantom

1. The problem statement, all variables and given/known data
A charge q1 = +8.15 muC is at the origin, and a charge q2 = -3.70 muC is on the x-axis 0.52 m from the origin. Find the electric field strength at point P, which is on the y-axis 0.61 m from the origin.
Answer = [1.613 x 10^5 N/C at 77.979 degrees]

2. Relevant equations
E = kQ/r^2
k = 9.0 x 10^9
tan o = Ey / Ex (?)
C^2 = A^2 + B^2 (?)
Ex = E1 cos 30 (?)
Ey = E2 - E1 sin 30 (?)

3. The attempt at a solution
I used the Pythagorean Theorem to figure out the distance from P to q2, which gave me 0.8 m.
I then went to use E to find E1, which I inputted as
((9x10^9)x(3.7x10^-6))/(0.8^2) = 5.2 x 10^4 N/C
Then E2 as
((9x10^9)x(8.15x10^-6))/(0.61^2) = 2.0 x 10^5 N/C
From there:
Ex = E1 x cos 30 = 4.5 x 10^4 N/C
Ey = (E2 - E1) x sin 30 = 7.4 x 10^4 N/C
Then I used Pythagorean Theorem again, thus getting:
E = Square root (4.5^2 + 7.4^2) x 10^4 = 8.7 x 10^4 N/C

Where did I go wrong? or did I do it incorrectly?

Last edited: Feb 12, 2007
2. Feb 11, 2007

### antonantal

I don't know where that "cos 30" came from, but you could simply add the vectors E1 and E2.

If $$a$$ and $$b$$ are two vectors with an angle $$C$$ between them, then their sum, $$c$$ can be calculated with this formula:

$$c^2=a^2+b^2-2abcosC$$

3. Feb 11, 2007

### FSNxPhantom

When i used that, I put in:

c^2 = (5.2 x 10^4)^2 + (2.0 x 10^5) - (2 x (5.2 x 10^4) x (2.0 x 10^5) x (cos 49.46)) = 1.7 x 10^5
I found 49.46 by:
Cos-1 (.52/.8) = 49.46

Still confused since it was wrong. Did i do it wrong? And if so, how do i found the angle for it being the answer:
[1.613 x 10^5 N/C at 77.979 degrees]

4. Feb 12, 2007

### antonantal

Take in account that the E field generated by a positive charge points away from it and the E field generated by a negative charge points towards it. Make a drawing and see what is the angle between E1 and E2, considering their directions.

5. Feb 12, 2007

### FSNxPhantom

P
l\
l..\
l....\
l......\
l_____\
q1->...<-q2

Is this it? Because I missed the class lesson and have to figure it all out right now.
And sorry for the bother since I'm clueless right now.

Last edited: Feb 12, 2007
6. Feb 12, 2007

### antonantal

You need to represent the E field in the point P, so draw E1 and E2 at P.

7. Feb 12, 2007

### FSNxPhantom

SOOOo....
E2______E
l.........../
l........./
l......./
l...../
l.../
l./
l/\ Angle
P_\_______
l\ E1./
l..\./ Angle from Cos o = (.52/.8) (?)
l....\
l......\
l........\
q1->..<-q2

Where the magnitude of E1 = 5.2 x 10^4 N/C and E2 = 2.0 x 10^5
Am I doing it right so far?

8. Feb 12, 2007

### antonantal

You represented correctly E1 and E2. E1 as you calculated it is the field generated by q2 and E2 is the field generated by q1, at point P.
You didn't take the right angle though. You need the angle between E1 and E2.

9. Feb 12, 2007

### FSNxPhantom

E2______E
l.........../
l........./
l......./
l...../
l.../
l./
l/\ Angle = o1 + o2 = 90
P_\_______
l\ E1./
l..\./ Angle from Cos o = (.52/.8) (?)
l....\
l......\
l........\
q1->..<-q2

And the angle between E1 and E2 is:
o = Cos^-1(.52/.8) + 90 degrees (from E2 to line in middle) ?

And for Ex and Ey (if there is)
Ex = E1 cos 49.46
Ey = E2 - E1 sin 49.46

Last edited: Feb 12, 2007
10. Feb 12, 2007

### antonantal

Yes. So the angle is 90 + arccos(0.52/0.80).
Now apply the formula that I gave you and, if the calculations are right you should get the right result for the magnitude of the E field at point P. You don't need Ex and Ey.

11. Feb 12, 2007

### FSNxPhantom

Angle is 90 + arccos(0.52/0.80) = 139.46
Then: a = (5.2 x 10^4) b = (2.0 x 10^5)
c^2 = a^2 + b^2 - 2abcos(139.46) = 2.4 x 10^5
c^2 = a^2 + b^2 + 2abcos(139.46) = 1.64 x 10^5 (close enough lol)

So how do I find the second part to the answer? [77.979 degrees]

12. Feb 12, 2007

### antonantal

To find the angle you must know one component of E, either Ex or Ey. Since to the component of E in the x direction only E1 has a contribution, it means that Ex is easier to get. So calculate the component of E1 in the x direction, E1x. This will be equal to Ex. Then, knowing Ex and E you can find the angle between them, right?

Last edited: Feb 12, 2007
13. Feb 12, 2007

### FSNxPhantom

Ex = E1 cos (49.46) = 3.38 x 10^4
E = 1.64 x 10^ 5
So:
Angle = arccos (Ex/E) = 78.1 degrees

14. Feb 12, 2007

### antonantal

Pretty close. There must be some approximations differences between your calculations and their.

BTW that formula, indeed should have been with a + . My mistake.

15. Feb 12, 2007

### FSNxPhantom

Thank you for your help. I'll see what my professor did to get that answer.