What is the magnitude E of the field?

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SUMMARY

The discussion focuses on calculating the magnitude of the electric field (E) affecting two charged spheres with equal but opposite charges of 72.0 nC and masses of 7.20 mg, suspended at an angle of 50.0 degrees due to the field. The user initially calculated E using incorrect angles and equations, leading to an erroneous result of 1.17 x 10^6 N/C. After correcting the angle used in calculations, the user arrived at 4.6 x 10^5 N/C, but this value was still incorrect, indicating further errors in the application of equilibrium equations.

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Homework Statement




Two tiny spheres of mass = 7.20 mg carry charges of equal magnitude, 72.0 nC , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m . When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to 50.0 degrees.

What is the magnitude E of the field?


Homework Equations


Fe1 = Eq
Fe2 = (Kq1q2)/r2
Fg = mg



The Attempt at a Solution



I assumed that the state it is in is in equilibrium so I set up:

ƩF = Fg + Fe1 + Fe2 + T = 0

ƩTx = Tcosθ - Fe1 + Fe2 = 0

θ being 65 degrees since I split the triangle made by the 50 degrees into 2 right triangles.

ƩTy = Tsinθ - mg = 0

T = (Fe1 - Fe2)/cosθ

Ty = (Fe1 - Fe2)tanθ - mg = 0

Fe1 = mg/tanθ + Fe2

E = (mg/tanθ + kq1q2/r2)/q

(I found r to be .448 using the right triangles I formed)

I plugged in all the given values and I've got...1.17 x 10^6

Which is wrong. Can someone please explain what I messed up on?
 
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Whoops I used the wrong angle during my calculations so my actual answer is 4.6 x 10^5 but it is still wrong
 

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