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MSCD Physics Department
2331002
Chapter 21 HW - grdhw
Time of Login: Thu Feb 5 13:46:16 2009
Deadline: Wed Feb 4 23:59:00 2009

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(DENN) 21.

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(21.
tanθ≈sinθ=(d/2L)

ΣF_x=TSinθ-F_repulsive = 0
ΣF_y=TCosθ-mg = 0

(mgsinθ/cosθ)=F_repulsive=(kq²/d²)

d³=(2kq²L/mg)

d alone equals (q²L/2πε₀mg)^1/3
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) no quiz

) 21.
After drawing the free body diagram I found that:
Tx = Tsinθ
Ty = Tcosθ

The force on x-axis:
Tsinθ = kq²/d^{2

The force on y-axis:

Tcosθ= mg

form the drawing:

d= 2Lsinθ

tanθ= kq2/mgd2

d=2Lθ according to small angle thearem.

d= 2Lq2/4πE0mgd2

d= (Lq2/2πE0mg2)1/3}^{Essay Grade: / Comment:}

21.
F = mg sinθ / cosθ
and
F = kq²/d²
tanθ = d/2L
so
F = kq² mg tanθ/d²
then solve for d.
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() 21.

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(21.

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(21.
from lecture:
k = 1/(4πε₀)
sinθ ~ tanθ ~ θ
θ = (d/2L)
tanθ = (kq²)/(mgd²)
(d/2L) = (kq²)/(mgd²)
d = ([kq²2L]/[mg])^1/3
d = ([q²L]/[mg2πε₀])^1/3

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(21.

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21.
d = (q2L/2ðå0mg)^1/3
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) 21.
= mgqTcos
εo) X (q²/d²)p= 1/(4qTsin
εod²mg)p= q²/(4qTan
≈sin≈d/2lqTan
εod²mg)pd/2l = q²/(4
d = (q2L/2πε0mg)^(1/3)

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(21.
T sin θ - k q² =0
T sin - mg = 0

sin θ = (d/2L)

tan θ = k q² / d^{2

d3 = k q2 2L / mg

d=[(q2 L)/ (2 π ε}₀^{m g)]}^{Essay Grade: / Comment:}

(21.
T_x = qE and T_y = mg
Tsinθ = kq²/d² and T = mg/cosθ
Substitute T: tanθ = kq²/d²mg
tanθ ≈ sinθ when θ is small and sinθ = d/2L, therefore:
d/2L = kq²/d²mg
d³ = 2kq²L/mg
d = (q²L/2πε₀mg)^1/3 QED

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(21.

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) 21.
ΣT_x=T sinθ-F_e=0
ΣF_y=T Cosθ-mg=0
T Sinθ=F_e
T Cosθ=mg
T=mg/Cosθ
mg*Sinθ/cosθ= F_e
Sinθ=d/2L
F_e=k*q²/d²
mgd/Cosθ*2L=k*q²/d²
d³mg=k*q²Cosθ2L
d³=2Lkq² Cosθ/mg
because tanθ≈Sinθ ;So Cosθ≈1
d=(q²*L/ 2πε₀mg)^1/3

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) 21.
ΣF_x=Tsinθ-F_e=0
ΣF_y=Tcosθ-mg=0

mgsinθ/cosθ=F_e=kq²/d²

d³=2kq²L/mg

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) 21.
sinθ=d/2L
tanθ=E/mg
d/2L=k*(q^2/d^2)
d^3=(q^2*L)/(2πεomg)
d = (q2L/2πε0mg)^1/3
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(21.
d = (q2L/2ðå0mg)^1/3
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(21.
Sinθ≈Tanθ
ΣF_x= Tsinθ-F_e=0
ΣF_y= Tcosθ-mg=0
F_e= mgSinθ/Cosθ = Kq²/d²
d³= 2Kq²L/mg
d=(q²L/2Piε₀mg)^1/3

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) 21.
Tcosθ=mg
Tsinθ=9E9(q²/d²)
tanθ=9E9(q²/d²)(1/mg)

tan= d/2L
d/2L=q²/4πε₀(d²)(mg)
d=((q²l/(2πε₀mg))^1/3

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21.

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() 21.

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(21.

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() 21.
sinθ=(1/2)d/L ; d=r ; q₁=q₂=q

F=(1/4πε₀)(q₁q₂/r²)

And F=mgsinθ

mgsinθ=(1/4πε₀)(q²/d²)

(1/2)d³=(1/4πε₀)(q²)(L/mg)

d=(q²L/2πε₀mg)^1/3

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(21.
sinθ=(1/2)d/L, d=r, q₁=q₂=q

F=(1/4πε₀)(q₁q₂/r²)

F=mgsinθ

mgsinθ=(1/4πε₀)(q²/d²)

mgd/2L=(1/4πε₀)(q²/d²)

d³=(1/2πε₀)(q²/mg)(L)

d=(q²L/2πε₀mg)^1/3

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(21.
è=arctan(qE/mg)
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) 21.

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21.
ƒ°Fx=TsinƒÆ-F=0 ¨ F=TsinƒÆ
ƒ°Fy=TcosƒÆ-mg=0 ¨ T=(mg)/(TcosƒÆ)
F=(kq2)/(d2)
sinƒÆ = tanƒÆ = (d)/(2L)
Therefore
F=(mgsinƒÆ)/(cosƒÆ) ¨
(kq2)/(d2) = mgtanƒÆ ¨
(kq2)/(d2) = (mgd/2L) ¨
d³ = (2kq2L)/(mg) ¨
d = [(q2L)/(2ƒÎƒÃomg)]1/3
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21.

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) 21.
d³= 2kLq²/mg
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) 21.
tanθ≈sinθ=(d/2L)

ΣF_x=TSinθ-F_repulsive=0
ΣF_y=TCosθ-mg=0

(mgsinθ/cosθ)=F_repulsive=(kq²/d²)

d³=(2kq²L/mg)

d alone equals (q²L/2πε₀mg)^1/3
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() 21.

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(21.
In the x-direction, the sum of the forces = Tsinθ-F_e=0
In the y-direction, the sum of the forces = Tcosθ-mg=0
So, mgsinθ/cosθ = F_e = kq²/d²
Solve for d: d³= 2kq²L/mg ===> d=(q²L/2πε⁰mg)^1/3

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(21.
Σ F_y = F_TCosθ - mg
F_TCosθ = mg

Σ F_x = F_TSinθ - 1/(2πε₀)*(q²/d²)
F_TSinθ = 1/(2πε₀)*(q²/d²)

Tanθ = Sinθ/Cosθ = q²/(2πε₀d²mg)

assume θ small so
tanθ ≈ Sinθ = d/(2L)

q²/(2πε₀d²mg) = d/(2L)
d = (Lq²/(2πε₀mg))^1/3

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(21.
If the two spheres are in equilibrium them the net force acting on them is zero. For this to be met The tension in the rope mus be balanced out by the force of mg down and the force of E to the right and left. So when θ is small sinθ and tanθ will get close to d/(2l,) so the force by E will approach dmg/(2L) and that will equall the force produced by E which is kq²/d². Then when you substitute 1/(4πε_o) in for k and solve for d you are left with d=(q²L/2πε_omg)^1/3
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21.
F = Eq
ΣFx = 0
Eq - Tsin(θ) = 0 equation (1)
ΣFy = 0
Tcos(θ) - mg = 0
T = mg/cos(θ) ---> plug into equation (1)

Eq = tan(θ)*mg

kq²/d² = tan(θ)*mg

small angle formula: tan(θ) ≈ sin(θ) ; sin(θ) = d/2L

q²/4πε₀ = d³mg/2L

d³ = q²L/2πε₀mg

d = (q²L/2πε₀mg)^1/3
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) 21.

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(SCHNEIDEA2156) 21.

Essay Grade: / Comment:

21.
tanθ≈sinθ=(d/2L)

ΣF_x=TSinθ-F_repulsive=0

ΣF_y=TCosθ-mg=0

(mgsinθ/cosθ)=F_repulsive=(kq²/d²)

d³=(2kq²L/mg)

d alone equals (q²L/2πε₀mg)^1/3
Essay Grade: / Comment:

) 21.
tanθ = kq²/mgd²
= d/2L = kq²/mgd²
= d³ = 2kq²L/mg
= d = (q²L/2πε₀mg)^1/3
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) 21.
I'll admit I have no idea on this one.
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(21.
F_x=T*sinθ-(q²/(4πε₀d²))=0
F_y=T*cosθ-mg=0
T is the tension force of the string. I've tried many different methods and I can't seem to solve this problem.

Essay Grade: / Comment:

21.
The System is at equilibrium therefore:
T_x=qE=Tsinθ=kq²/d²
T_y=Tcosθ=mg
sinθ=d/2L
Tsinθ/Tcosθ=tanθ=kq²/mgd²
tanθ≈sinθ (small angle formula)
tanθ=sinθ=kq²/mgd²=d/2L
d³=2Lkq²/mg
d=(q²L/2πε₀mg)^1/3^{Essay Grade: / Comment:}

(21.
d=√((q²L)/(4πmgqEε₀))
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) 21.
tanθ = (1q²/4πε₀d²)(1/mg)
also tanθ = d/2L
d³ = Lq²/2πε₀mg
solve for d = (Lq²/2πε₀mg)^1/3

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) 21.
At equilibrium:
T cosθ = mg
T sinθ = (1/4πε₀)(q²/d²)
tanθ = q²/ 4πε₀d²(mg)

tanθ ≈ d/dl
d/2d = q²/ 4πε₀d²(mg)

Giving you the seperation between the sphere is (d)= [q²/ 2πε₀d²(mg)]^1/3
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(21.
Fe=Tsintheta=q1q2/d²4piepsilon

T=mg/costheta

mg sintheta/costheta=q1q2/d²4piepsilon

mgtantheta=q²d²4piepsilon

mgsintheta=q²d²4piepsilon

mgd/L=q²d²4piepsilon

d²/d=q^{224piepsilonmg

d=(q2L/4piepsilonmg)1/3}^{Essay Grade: / Comment:}

() 21.
ΣFx = 0
Tx - q E = 0
Tx = T sinθ
T sinθ - q E = 0
T sinθ = (K q²)/d²
T = [(K q²)/d²] X (1/sinθ)

ΣFy = 0
Ty - m g = 0
Ty = T cosθ
T cosθ = m g
T = (m g) / cosθ

(K q²)/d² = (m g sinθ)/cosθ

tanθ ≈ sinθ = d/(2L)

d³ = (2 K q² L)/(m g)

d = [( q² L)/(2 π ε_o m g)]^1/3

d increases when q increases

Essay Grade: / Comment:

(21.
From the free body diagram:
T_x = qE
T_y = mg
Therefore, Tsinθ = qE = kq²/d²
Tcosθ = mg
Tsinθ/Tcosθ = kq²/d²mg
tanθ = kq²/d²mg
Since θ is small then tanθ≈sinθ≈θ
therefore, θ = kq²/d²mg
also θ = sinθ = d/2L
So, d/2L = kq²/d²mg
Since k = 1/4πε₀
d = (q²L/2πε₀mg)^1/3
Essay Grade: / Comment:

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