Electric Field and Force Components for Point Charges Q1 and Q2?

[formulajoe]

point charges, Q1 (4,0,-3) and Q2 (2,0,1). Q2 = 4 nC, find Q1 such that the a. electric field at (5,0,6) has no z component.
b. force on a test charge at (5,0,6) has no x component.

heres what i did for a:

i found the distance from Q1 to (5,0,6) to be the square root of 82
distance from Q2 to (5,0,6) to be square root of 34.

i than found the unit vectors.
multipled by the charges and set the z components equal.

so i had 9Q1/sqr rt 82 = - 4nc*5/sqr rt 34
ended up with Q1 = -3.41 nc. answer in the back of the book says -3.463

on the second part i found the two forces using (Q*Qt)/4*pi*Eo*R^2.
i than multiplied these by the unit vectors in the x direction and set one of them equal to negative the other and solved for Q1. i ended up with an answer of -48nC, the book says -18nC.

 

[dextercioby]

Why are there square roots in the denominators in the first equality.Shouldn't they have disappeared when squared...?

Daniel.

 

[formulajoe]

thats how the unit vector is found. divide the change in x or y or z and than divide by the distance between the two points. distance between the two points is the square root of (x2-x1) + (y2-y1) + (z2-y1)

 

[dextercioby]

Yes,but you should have equated "z" components of the fields ALTOGETHER...Does that change anything...?

Daniel.

 

[formulajoe]

so i find electric field due to Q1 and Q2 by using this formula right:
E = F/Q
where F = (Q*Qt)/(4*pi*Eo*R^2). i don't see how that's going to work because the Qs are going to cancel.

 

[dextercioby]

No,they're not.What's the general expression of an electric field generated by a point charge "q" situated in the origin of a coordinate system,field created at point of position vector \vec{r} ??

Daniel.

 

[formulajoe]

E = (q*Q/4*pi*Eo*R^2)/q Ar
right?

 

[dextercioby]

No.It's more like a vector,of this form
\vec{E}=\frac{q}{4\pi\epsilon_{0}|\vec{r}|^{2}} \frac{\vec{r}}{|\vec{r}|}

For your problem,chose a system of coordinates and compute the total field at the point you're interested.Just then,u can set the component of the resulting field equal to zero.

Daniel.

 

[formulajoe]

k, what about the second part? I am pretty sure i did that right

 

[formulajoe]

and i tred it the way you told me and i got an answer of -8.3 nC for Q1. not the same as the back of the book.

 

[dextercioby]

For the second part,you need to have done first part right...Because it uses the charge calculated at point "a"...

Daniel.