# Charge within an electric field

1. Feb 3, 2010

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

A -10.0 nC charge is located at position (x,y) = (2.0cm, 1.0cm). At what (x,y) position(s) is the electric field:

161,000i - 80,500j N/C

x = ??? cm
y = ??? cm

2. Relevant equations

E = Kq/r2

3. The attempt at a solution

I know that the magnitude of E is $$\sqrt{161,000^{2}+80,500^{2}}$$ = 1.8e5.

So, to find the distance between the electric charge and where the field is:

1.8e5 = Kq/r2

Therefore, r = $$\sqrt{(9e^{9})(10e^{-9})/1.8e^{5}}$$ = .02m = 2cm

So, I know the distance between the charge and the electric field. The angle that the electric field makes with the x-axis is tan-1(-80,500/161,000) = -26.57

I was thinking that I have to do something with polar coordinates to finish the problem but I can't seem to make the mental leap.

I know in polar coordinates, if you're given a distance (2cm) and an angle (-26.57), you should be able to find the rectangular coordinates by:

x = 2cos(-26.57)
y = 2sin(-26.57)

But isn't this the position of the electric field in relation to the origin (not the charge)?

2. Feb 3, 2010

### Delphi51

I played with this a bit and may have found an insight. First, I noticed that for the (x,y) answer, x > 2 (because the Ex is positive) and y > 1 because Ey is negative. I sketched the vector and saw that at (x,y) the E field is
kq/r²[(x-2)/r*i - (y-1)/r*j]
If you write that the x component is 161000 and the y component is -80500 you get a pair of equations that can be partly solved very quickly. Solve one for kq/r² and sub that into the other. You get the equation of a straight line on the x,y plane. Combine that with your value for the radius and you'll soon have the point on that line with the right radius and x > 2, y > 1.

3. Feb 3, 2010