# E Field Question

1. Jan 31, 2008

### And 1

1. The problem statement, all variables and given/known data
Point charge Q1 = 25nC located at P1(4,-2,7) and a charge Q2 = 60 nC be at P2(-3,4,-2). At what point on the y-axis is Ex=0?

2. Relevant equations
E= [ Q / (4*pi*E0*r*r) ] ar

3. The attempt at a solution

I took:
( Q1 / 4*pi*E0*(x*x+y*y+z*z) * (y / sqrt(x*x+y*y+z*z)) ay

I got -6.51.

The solution is -6.89.

Am I way off base or on the right path? Thanks for your help.

2. Jan 31, 2008

### And 1

It appears I might have been way off base. Reworking a different thought...call off the Help Department (for now). :)

Report back in a few...

3. Jan 31, 2008

### And 1

Ok. I'm stuck again...
Basically we want Ex=0 at some point (P3) on the y-axis. So the point P3 (0,y,0).

Taking the two points to get the total E field. ET = E13 + E23.

I've got:

E13 = 25nC [(0-4)ax + (y+2)ay + (0-7)az ] / (4*pi* 8.854e-12 [(-4)^2 + (y+2)^2 + (-7)^2]^3/2)

E23 = 60nC [(0-(-3)ax + (y-4)ay + (0+2)az ] / (4*pi* 8.854e-12 [(3)^2 + (y-4)^2 + (2)^2]^3/2)

Kind of stuck at this point...any help? Thanks.

Last edited: Jan 31, 2008
4. Feb 22, 2008

### unplebeian

I saw throught your work very quickly. Here's what I have to offer:

Now since you want Ex= 0 equate the x components of E13 and E23 together and take their magnitude. Only when they are EQUAL and OPPOSITE in magnitude and direction respectively, will you get Ex= 0.

5. Feb 22, 2008

### unplebeian

Yes, that works! You can do it in two ways. From the above post, you can either calculate the value for y, or you can plug in your answer y= -6.89 from the solutions manual that you have and find out that both the magnitudes are the same.

So

25nC* -4
----------------------------- = -0.11927*10^-9 when y= -6.89
[16+ (y+2)^2 + 49 ]^3/2

and

60nC* 3
----------------------------- = -0.11924*10^-9 when y= -6.89
[9+ (y-4)^2 + 4 ]^3/2

I think that's pretty close enough!

I hope that helped.

6. Feb 22, 2008

### And 1

Yes, it did. Thanks for your help.