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What is the magnitude E of the field?

  • Thread starter phyxstdt
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
2
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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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