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the forces in each direction:
∑ Fx = 0
F T sin 20° − FE = 0
FT = FE
sin 20° ∑ Fy = 0
F T cos 20° − Fg = 0
FT = Fg
cos 20° Since we have two equations (and we only want one), we are probably going to try
combine these equations somehow. Since both equations have FT, we can isolate this
variable and then equate the two equations. Additionally, we will introduce the equation
for Coulomb’s Law for the variable FE and the regular mg term for Fg. Solutions will be posted at www.prep101.com/solutions 4 ©Prep101 Phys1026 April Exam Solutions FT = FT
Fg
FE
=
sin 20° cos 20°
sin 20°
FE = Fg
cos 20°
= Fg tan 20°
FE
q1 q 2
= mg tan 20°
r2
k c q1 q 2
r2 =
mg tan 20° kc r= k c q1 q 2
mg tan 20° plugging in q1 = 2.0 µC, m = 50g and q2 = 3.0 µC and remembering to convert the mass
to kg:
r = 0.55m
Question 2:
Solution:
We have to deal with the entire question algebraically. All of the charges are positive, so
they repel each other. Each charge is repelled by two other charges. If we remove one
charge, then the amount of repulsion will decrease, so the force on each charge will also
decrease. Again choose our x and yaxes: the xaxis is along the horizontal and positive
to the right and yaxis is along the vertical and positive in the upward direction. Now we
will create a FBD for one of the charges: The net force F0 on a single charge would be equal to the vector sum of two equal forces
(each is FE) separated by an angle of 60°. To get the net force, which is the original force
on each charge F0 determine the net force in each of the axis directions and then find the
net total force: Solutions will be posted at www.prep101.com/solutions 5 ©Prep101 Phys1026 April Exam Solutions Fx = − FE − FE cos 60° Fy = − FE sin 60° Fx = −1.5FE Fy = −0.866 FE Fnet = F0 = Fx2 + Fy2
F0 = (− 1.5FE )2 + (− 0.866 FE )2 F0 = 1.73FE
Once one of the charges is removed, the force decreases from 1.73 FE to FE. The new
force is less than the original force by a factor of 1/1.73. The new force on each charge is
FE = 0.58 F0.
2.3 Answer (B) The flight time through the plates in the xdirection is given by v = L / t ⇔ t = L /v and
we have F = qE = may ⇔ ay = qE / m. If we set the initial position of the droplet in the
ydirection to zero then the maximal allowed displacement is d/2. We also have that the
initial velocity in the ydirection is zero. Thus we have:
sy(t) = 0.5⋅ ay⋅ t2
⇔
⇔ d/2 = 0.5⋅ q⋅ E⋅ L2 / (m⋅v2)
q = m⋅ v2⋅ d / (E⋅ L2) 2.4 Solution:
(a) Use Coulomb’s Law to find the force of the charge at B on the charge at A:
r
kq q
k (+Q)(−Q) kQ 2
kQ 2
ˆ
FBonA = a2 b =
= 2 [RIGHT]
→
FBonA = 2 x
rab
(d ) 2
d
d
r
kq q
k (+Q)(−Q ) kQ 2
kQ 2
ˆ
FConA = a2 c =
= 2 [UP]
→
FConA = 2 y
rac
(d ) 2
d
d
The net force on the charge at A is the vector sum of these two component forces:
r
⎛ kQ 2 ⎞ ⎛ kQ 2 ⎞ ⎛ kQ 2 kQ 2 ⎞
Fnet = ⎜ 2 ,0 ⎟ + ⎜ 0, 2 ⎟ = ⎜ 2 , 2 ⎟
⎟ ⎜ d ⎟ ⎜d
⎜d
Fnet
d⎟
FConA
⎠
⎠⎝
⎠⎝
⎝
Fnet ⎛ kQ...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

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