Finding the Distance for a Zero Electric Field at a Corner of a Square

[thercias]

Homework Statement

Two tiny objects with equal charges of 63.0 µC are placed at two corners of a square with sides of 0.260 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?

Homework Equations

E = KQ/r^2

The Attempt at a Solution

I first solved for E net on point A. My values were
Enetx = 2.962 x 10^6
Enety = 1.13 x 10^7
Magnitude of Etotal = 1.17 x 10^7

angle formed : 75.3 degrees

Now I tried two alternate solutions, but in the end I got the wrong answer still for both tries.
E = kq/r^2
r = sqrt(kq/E)
r = sqrt(k(63e-6)/1.17e6)
r=0.22m

dx=0.22cos75.3=0.0558
dy=0.22sin75.3=0.213

That one was wrong, so I tried finding the value of r using the components of the electric field
rx= sqrt(k(63e-6)/2.962e6)
ry=sqrt(k(63e-6)/1.13e7)

rx =0.43m
ry=0.22m

still wrong, i need help because I'm not sure what I'm doing wrong.

 

[Simon Bridge]

The first method is the one I'd have used.
Check units and arithmetic.

I first solved for E net on point A. My values were
Enetx = 2.962 x 10^6
Enety = 1.13 x 10^7
Magnitude of Etotal = 1.17 x 10^7

angle formed : 75.3 degrees

... that angle looks bad - the resulting field should point up and to the left, so the angle from the +x axis will be bigger than 90deg.

I don't see how you got the x and y components either.

 

[thercias]

The first method is the one I'd have used.
Check units and arithmetic.

I first solved for E net on point A. My values were
Enetx = 2.962 x 10^6
Enety = 1.13 x 10^7
Magnitude of Etotal = 1.17 x 10^7

angle formed : 75.3 degrees

... that angle looks bad - the resulting field should point up and to the left, so the angle from the +x axis will be bigger than 90deg.

I don't see how you got the x and y components either.

I ussed the equation

E1 = kq/r^2
= kq/0.26^2
= 8.378e6 going south

E2 = kq/(0.26^2+0.26^2)
= 4.189e6 going south and to the right
E2cos45 = 2.96e6 (to the right)
E2sin45 = 2.96e6 (south)
therefore Eynet = 1.13e7
Exnet = 2.96e6
Magnitude sqrt ((2.962e6)^2+(1.13e7)^2)
=1.17e7
angle = tan inverse (Ey/Ex)
= 75.3 degrees

So the electric field will go in the opposite direction of this to cancel out of the third charge.
I might be doing something wrong, but I don't know what it is. This is due in 42 minutes, if anyone can check my values quickly or if there's a logic error that I missed i'd appreciate it.

 

[Simon Bridge]

How do you kow it's wrong - is it computer moderated?
If so - careful with rounding - make sure you have entered the correct format and units.
i.e. does it want an angle? If so - in degrees or radians?
How many sig fig is it expecting?

You realize that 75.361° is clockwise from the -x axis - if they want to measure angles anticlockwise from the +x then this will be marked wrong. Also notice that it rounds to 75.4 not 75.3.

You got to check the details.

I'm getting the same angle and field strength as you, but a different distance, using k=8.99e9(SIU)

 

[Simon Bridge]

[edit]nope - I just forgot to invert.
I'm getting the same as you - check for fiddly stuff.

 

[thercias]

edit: nevermind, thanks simon

 

[thercias]

How do you kow it's wrong - is it computer moderated?
If so - careful with rounding - make sure you have entered the correct format and units.
i.e. does it want an angle? If so - in degrees or radians?
How many sig fig is it expecting?

You realize that 75.361° is clockwise from the -x axis - if they want to measure angles anticlockwise from the +x then this will be marked wrong. Also notice that it rounds to 75.4 not 75.3.

You got to check the details.

I'm getting the same angle and field strength as you, but a different distance, using k=8.99e9(SIU)

Ok this helped me out. I guess I just wasn't being detailed with my calculations, it was off by 0.02 and that was enough to make it wrong on the computer. Thanks so much!