https://padlet.com/tec430/oq19wxyacm7p (Problems 6-8 attatched in Schoology) en-us 2017-04-25 15:21:15 UTC 2017-04-28 19:13:06 UTC hello@padlet.com https://padlet.com/tec430/oq19wxyacm7p/wish/168667672 6. The first arrangement has the greatest net force because the charges on either side of the center alternate. The other two arrangements have a net force of zero because the charges of the particles on either side of the central charge are equal. Thus, the direction of the net force if the central charge is positive, is to the right.
7. F1=[k(q)(2q)]/.0025=.07192 N
F2=[k(2q)(-2q)]/.0025= -.14384 N
F3=[k(2q)(-q)]/.005= -.03596 N


]]> 2017-04-27 15:29:21 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168667672 https://padlet.com/tec430/oq19wxyacm7p/wish/168667865 6. Arrangement 1 will have the largest net force acting upon it. Since the other arrangements have the same arrangement on either side of the center dot in respect to the dot, they will have the same effects, and therefore, cancel each other out. The direction of the net force of arrangement 1 would be to the right, since it is repelled from the positive charge on the left, and attracted to the negative charge on the right. 
7. Interaction with upper-right:
F=k(2.0x10^-7C)(-1.0x10^-7C)/(0.005m^2)
F=-0.036N
Interaction with bottom-right:
F=k(2.0x10^-7C)(-2.0x10^-7C)/(0.0025m^2)
F=-0.14N
Interaction with upper-left:
F=k(2.0x10^-7C)(1.0x10^-7C)/(0.0025m^2)
F=0.072N
Vector addition:
(.036)sin(45)=0.025
(.036)cos(45)=0.025
x=0.025+0.14
x=0.16
y=0.025-0.072
y=-0.047
Tan(-0.047/0.16)
-.0051 degrees
(0.16^2+0.047^2)^-.5
0.17N
8. ]]> 2017-04-27 15:29:48 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168667865 https://padlet.com/tec430/oq19wxyacm7p/wish/168667984 6)Arrangement 1 will have the largest net forces because the charges in the first arrangement are not in the same position on each side. 
7)q=1.0 x 10^-7   K=8.99x10^9
F=(8.99 x10^9)(1.0 x 10^-7)/(0.05)^2 
F=359600
F2=]]> 2017-04-27 15:30:09 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168667984 https://padlet.com/tec430/oq19wxyacm7p/wish/168670063 6. The first row will have the greatest net force acting upon it because the charges of the opposite sides interchange. The other two rows will have a 0 net force because the charges cancel each other out. The direction of the net force is to the right because the central charge is positive
7. 
k=8.99 x 10^9, q= 1.0 x 10-7C
F1=(k*q*2q)/(r^2)
F1=((8.99 x 10^9)(1.0 x 10^-7)(2* 1.0 x 10^-7)) /(.05^2) = .07192N
F2=(k*2q*-2q)/(r^2)
F2=((8.99 x 10^9)(2 x 1.0x10^-7)(-2 x 1.0 x 10^-7))/(r^2)= -.14384N
F3=(k*2q*-q)/(r^2)




]]> 2017-04-27 15:36:17 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168670063 https://padlet.com/tec430/oq19wxyacm7p/wish/168671437 6) The first arrangement has the greatest net force acting on it because positive charge pushes it away and toward the negative charge, so its  huge pull to the right side.

7) q = 1.0 x 10^-7
k = 8.988 x 10^9

F1 = (k)(q)(2q)/ (0.05^2) = (2k*q^2)/(0.05^2) = 7.19 x 10^-2 N

F2 = (k)(2q)(2q)/ (0.05^2)= ]]> 2017-04-27 15:40:09 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168671437 https://padlet.com/tec430/oq19wxyacm7p/wish/168671915 6) Arrangement 1 has the greatest net force acting on it since the force on the left is pulling away and the force on the right is pulling toward the center. Therefore the net force is pulling to the right. Meanwhile the other two net forces cancel each other out. 
7) k=9x10^9 q=1x10^-7
(k*q*2q)/(r^2)
(9*10^9)(1*10^-7)(2*10^-7)/.05^2)=.072N
(k*2Q*-2Q)(r^2)
(9X10^9)(2X10^-7)(-2x10^-7)/(.0025^2)=-.14N
(9*10^9)(2*10^-7)(1*10^-7)/(.0025m^2)=

]]> 2017-04-27 15:41:17 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168671915 https://padlet.com/tec430/oq19wxyacm7p/wish/168675284 6) The third central charge has the greatest net force acting on it because the same charges on each side repulse each other which applies force to the central charge. The opposite charges on each side attract each other so there isn't a force applied to the central charge. 
7) q= (1.0x10^-7)
k= (8.99x10^9)
F1= (k)(q)(2q)/(.05^)
(2k*q^2)/(.05^2)= 7.19x10^-2 N

F2= (k)(2q)(-2q)/(.05^2)=]]> 2017-04-27 15:50:49 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168675284 https://padlet.com/tec430/oq19wxyacm7p/wish/168676656 6)  The first arrangement has the greatest net force on it.  The other two arrangements have a net force of 0.  The center charge is forced 1 q to the right if it is positive.
7)q= (1.0x10^-7)
k= (8.99x10^9)
F1= [(2kxq^2)]/[(0.05^2)] = 7.19x10^-2N
F2= [(8.99x10^9)(2x(1.0x10^-7)(-2(1.0x10^-7)]/[(r^2)] = -1.44x10^-2N
F3= [k(2q)(-q)]/(0.05) = -3.59x10^-2N
8)m=(1.0*10^-6)
g=9.8m/s^2
r=0.25m
Fg=9.8((1.0*10^-6)^2)/(0.25)^2)]]> 2017-04-27 15:54:27 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168676656 https://padlet.com/tec430/oq19wxyacm7p/wish/168679981 1. The second and third line have a very small or zero net force because the central charge is either equally repelled or equally attracted. The first line would have the greatest net force, since the left side repels the central charge while the right side pulls the central charge toward the pull of the closest right charge.
7. 
F1= (8.988x10^9)(1.0x10^-7)((2)X(1.0x10^-7))/(0.05^2)

F2=(8.988x10^9)((2)X(1.0x10^-7))((-2)X(1.0x10^-7))/(0.05^2)

F3= ]]> 2017-04-27 16:05:12 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168679981 https://padlet.com/tec430/oq19wxyacm7p/wish/168682490 6.) The third central charge has the largest net force because it is made up of all the same charge. The first one has a net force of zero and the second has a positive charge in the middle. 
7.) F1 = ((8.99*10^9)(1.0*10^-7)(q2))/0.05^2
7.19 * 10 ^-2
F2= ((8.99*10^9)(1.0*10^-7)(1.0*10^-7))/(0.05^2)
-1.428*10^-6
F3=((8.99*10^9)(2q)(-1q))/0.05
-3.38*10^-2
8.) 4.24 * 10^-6 = ((8.99 * 10^9)(q1)(q2))/(0.25)^2
5.4*10^-9]]> 2017-04-27 16:12:30 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168682490 https://padlet.com/tec430/oq19wxyacm7p/wish/168682647 F2 = (k)(-4q^2) / (0.05^2) = -1.44 x 10^-1 N

r = (0.05^2 + 0.05^2)^(1/2) = 0.0707 m
F3 = (k)(2q)(-q) / r^2 = -3.60 x 10^-2 N

Ftot = F1 + F2 + F3 = -0.1081 N

8) ]]> 2017-04-27 16:13:02 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168682647 https://padlet.com/tec430/oq19wxyacm7p/wish/168763079 (9*10^9)(2*10^-7)(2*10^-7)(1*10^-7)/.05^2=.072N
8. 1x10^-6
r=.25m
4.24x10^-6N=9x10^9(q1)(q2)/.25^2
5.4x10^-9

]]> 2017-04-27 21:25:45 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168763079 https://padlet.com/tec430/oq19wxyacm7p/wish/168772205 6.) The first arrangement has the greatest force acting on it because the forces are not balanced on both sides. It is not symmetric. If the central particle was positive, the particle will move to the right on the first one. The second and third have no net force
7.) The vertical force is 2KQ^2/.0025, the horizontal force is -4KQ^2/.0025 and the diagonal force is -2KQ^2/.005, where K is Coulomb's Constant and Q 10^-7.
The X component is the sum of the horizontal and the diagonalXSqrt(2)/2, and the Y component is the sum of the vertical and the diagonalXsqrt(2)/2. If we square both the X and Y components and sum them, and then sqrt them, we apply Pythagoras's theorem and come out with .138 N of net force on the particle.
8.)]]> 2017-04-27 23:22:35 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168772205 https://padlet.com/tec430/oq19wxyacm7p/wish/168776787 6. Arrangement 1 because it is not symmetrical. There is an uneven balance when compared to 2 and 3, seeing as to how on either side it is equal in terms of what forces are acting on them.
7. q = 1.0* 10^-7 K= 8.99*10^9
F=(10^9*8.99)(1*10^-7)/(.05)^2
F= 359600

]]> 2017-04-28 00:18:46 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168776787 https://padlet.com/tec430/oq19wxyacm7p/wish/168784811 F1=.0719N
F2=-.143800N

F3= (8.988x10^9)(2)(1.0x10^-7)(-1)x(1.0x10^-7)/(0.5^2+0.5^2)^1/2=-.035952N

8. ]]> 2017-04-28 01:45:36 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168784811 https://padlet.com/tec430/oq19wxyacm7p/wish/168793386 7 (cont.)
F3= (k*2q*-q)/(r^2)
F3= ((8.99 x 10^9)(2 x 1.0 x 10^-7)(-1.0x10^-7))/(.05^2)= -.03596


]]> 2017-04-28 03:38:35 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168793386 https://padlet.com/tec430/oq19wxyacm7p/wish/168802351 7. Ftot= F1+F2+F3= -0.10788 N]]> 2017-04-28 05:57:14 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168802351 https://padlet.com/tec430/oq19wxyacm7p/wish/168915581 7) F2= (k)(-2q)(2q)/(.05)^2
=-1.44x10^-2N
F3= (k)(2q)(-q)/(.05)^2
=-3.59x10^-2N
8) m=(1.0x10^-6)
g=9.8m/s^2
r=.25m
Fg-= (9.8)(1.0x10^-6)^2/(.25)^2


]]> 2017-04-28 15:55:20 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168915581 https://padlet.com/tec430/oq19wxyacm7p/wish/168966460 F2=(8.99*10^9)(1.0*10^-7)
= -1.428*10^-6
F3= -3.38*10^-6 
8)]]> 2017-04-28 19:10:19 UTC https://padlet.com/tec430/oq19wxyacm7p/wish/168966460