Period 2 Coulumb's Law Practice 2 by Thomas E Clark

Arnav Verma

2017-04-25 16:20:26 UTC

1. According to Coloumb's Law, the top arrangement is the only one that will produce a net forcee on the central charge. The middle and the lower arrangement produce a net force of zero. If the central charge is positive, then the net force is towards the right.
2. k(2q)(q)/a^2 = 0.072 N
k(2q)(q)/((sqrt(2a))^2 = 0.036 N
k(2q)(2q)/a^2 = 0.144 N
0.036cos(45) + 0.144 = 0.17 N
0.144sin(45) - 0.072 = -0.05 N
F = 0.18 N
3, (1*10^-6)(9.8)/tan(60) = (9*10^9) (q^2)/(0.25^2)
q = 6.3 *10^-9

Tabib Hossain

2017-04-25 16:20:39 UTC

1. The first line has the greatest net force acting on it. The net force on the other two lines is 0. The direction of the net force on the first line is towards the right.
2. k(2q)(q)/a^2 = 0.072N
k(2q)(q)/((sqrt(2a))^2=0.036N
k(2q)(q)/a^2=0.144N
0.036cos450.144=0.17N
0.144sin45-0.072=-0.05N
F = 0.18N
3. (1*10^-6)(9.8)/tan60 = (9*10^9)(q^2)/(0.25^2)
q = 6.3*10^-9C

Arib

aahmed5 — 2017-04-25 16:21:36 UTC

1. In Coloumb's Law, the first arrangement will product the greatest net force on the central charge. The second force will produce a net force of zero also with the third one. If the central charge is positive, the net force is directed towards the top or right.

2. k x (q q) / a2 = 0.072 N
k (2q)/sqrt(2a) 2 = 0.036 N
k (2q)(2q) / a2 = 0.144 N
0.036 cos(45) + 0.144 = 0.17 N
0.144 sin 45 - 0.072 = -0.05 N
F = 0.18 N

3.(1 x 10^-6)(9.8)/tan 60 = 9 x 10^9 x q2 / 0.25^2

q = 6.3 x 10

Adithya Gutala

2017-04-25 16:21:48 UTC

1. The first diagram is the arrangement that will produce a net force on the central charge.

Matthew Aragaw

2017-04-25 16:33:51 UTC

Andrew Lin

alin3 — 2017-04-25 16:41:34 UTC

6. The first arrangement will produce the greatest net force on the central charge (in fact, it is the only arrangement that will produce a net charge at all). The central charge is attracted to one particle and repelled in the same direction by the particle on the opposite side. The second and third arrangements produce a net force of zero. If the central charge is positive, the net force is directed to the right (the charge is repelled from the like charge on the left and attracted to the opposite charge on the right).

7. Repulsion between UL and LL: F = k(q)(2q) / (0.050 m)² = 0.072 N at 270° = <0, -0.072 N>
Attraction between UR and LL: F = k(-q)(2q) / (0.050√2 m)² = 0.036 N at 45° = <0.025 N, 0.025 N>
Attraction between LL and LR: F = k(2q)(-2q) / (0.050 m)² = 0.14 N at 0° = <0.14 N, 0>
Net force on LL: <0.025 N + 0.14 N, -0.072 N + 0.025 N> = <0.17 N, -0.046 N> = 0.18 N at 345°

8. F_g = (1.0 × 10^-6 kg)(9.8 m/s²) = 9.8 × 10^-6 N at 180° = <0, -9.8 × 10^-6 N>
F_net = 0, F_n at 60°, so F_g + F_elec at 240°
F_elec at 180°, sum of gravitational and electrical forces on particle = elec, -9.8 × 10^-6 N>
tan240° = -9.8 × 10^-6 N / F_elec
F_elec = 5.7 × 10^-6 N at 180° = k(Q)² / d²cos 60° = ((d / 2) / 0.25 m)
d = 0.25 m
5.7 × 10^-6 N = k(Q)² / (0.25 m)²Q = √((5.7 × 10^-6 N)(0.25 m)²/ k) = 6.3 × 10^-9 C

Anton Loeb

aloeb1 — 2017-04-25 16:50:52 UTC

1. The first figure has the greatest net force being acted upon it. The Net force of the other lines is zero by the laws of electricity, and the net force is going in the right(positive) direction.
2. K92q)(q)/a^2 = .072 N
k(2q)(2q)/a^2 = .144 N
.036cos(45) + .144 N = .17 N
Force is .18 Newtons.
3. ((9*10^9) (q^2)/(.25^2)) = ((1 * 10^-6)(9.8)/(tan60))
Q charge is 6.3 * 10^-9 Coulombs.

Kayla Yim

kaylayim5 — 2017-04-26 15:24:21 UTC

1. The first diagram will produce the greatest net force on the central charge because the other diagrams do not have a net charge at all. The central charge is repulsed and attracted on either side.
2. K (2q^2)/ a^2= 0.072N
K (2q^2)/ (sqrt 2a)^2= 0.036N
K (4q^2)/ a^2= 0.144N
0.036cos(45) + 0.144= 0.17N
0.144sin(45)- 0.072= -0.05N
F= 0.18N
3. (1*10^-6)(9.8)/tan(60)=((9*10^9)(q^2))/ (.0625)
q= 6.3*10-9

Adithya Gutala

agutala — 2017-04-26 15:25:26 UTC

1. The first diagram is the arrangement that will produce a net force on the central charge.
The other two produce zero
2. k (2q^2)/a^2 = 0.072N
k(2q^2)/(sqrt(2a))^2 = 0.036 N
k(4q^2)/a^2 = 0.144 N
0.036 cos(45) + 0.144 = 0.17 N
0.144 sin(45) - 0.072 = -0.05 N
F = 0.18N
3. (1*10^-6)(9.8)/tan(60) = (9*10^9) (q^2) /(.0625)
q = 6.3 * 10^-9

Matthew Aragaw

maragaw — 2017-04-26 15:25:48 UTC

1. According to Coloumb's Law , the first arrangement will produce the greatest net force on the central charge .
2. k (2q^2)/a^2 = 0.072 N
k (2q^2)/(sqrt(2a))^2 = 0.036 N
k (4q^2)/a^2 = 0.144 N
0.036 cos(45) + 0.144 = 0.17 N
0.144 sin(45) - 0.072 = -0.05 N
F = 0.18
3. (1*10^-6)(9.8)/tan(60)

Alan Martino

2017-04-26 15:27:08 UTC

6.) The figure that comes in first seems to produce the largest net force on the central charge. The net force of the other lines produces zero.
7.) K(2Q) / (.050 M)^2 = .072 N @ 270 Degrees = (0,-0.072N)
The attraction b/w UR and LL: F = K(-Q)(2Q) / (.050 ROOT 2M)^2 = .036 N at 45 degrees = (0, -.072 n) Attraction b/w LL AND LR: F= K(2Q) )-2Q) / (.050 M)^2 = .14 N at 0 degrees = (.14 N, 0)
Net force on LL: (0.025 N + .14 N, -0.072 + .025 N) = (.17 N, -0.046 N) = .18 N at 345 degrees
8. Fg = )1.0 x 10-6 Kg) (9.8 m/s2) = 9.8 x 10^-6 N at 180 = (0, -9.8 x 10^-6N)

Weixin Luo

2017-04-26 15:27:23 UTC

1. The top central charge has the greatest net force acting on it. The other 2 produce 0 net force because the positive net force will be towards the right.
2.K(2q^2)/a^2=0.072N
K(2q^2)/(sq(2a))^2=0.036N
K(4q^2)/a^2=0.144N
0.036cos45+0.144=0.17N
0.144sin45-0.072=-0.05N
0.17+0.05=~0.18N
3. Fg=(1.0x10^-6)(9.8m/s^2)/tan60=6.3x10^-9 c

Stephen Tu

2017-04-26 15:39:18 UTC

6. The first arrangement produces the largest net force on the central charge.
7. k (2q^2)/a^2 = 0.072 N
k (2q^2)/(sqrt(2a))^2 = 0.036 N
k (4q^2)/a^2 = 0.144 N
0.036 cos(45) + 0.144 = 0.17 N
0.144 sin(45) - 0.072 = -0.05 N
F = 0.18
8.
3.(1 x 10^-6)(9.8)/tan 60 = 9 x 10^9 x q2 / 0.25^2
q = 6.3 x 10^-9

Natalie Pandher

2017-04-26 15:45:56 UTC

6. The first central charge has the largest net force acting on it. The second point has no net force because the positive charges repel each other and do not act on the center point.

7. k*(2*q^2)/(a^2) = 0.072 N
k*(2*q^2)/(sqrt(2a))^2 = 0.036 N
k*(4*q^2)/(a^2) = 0.144 N
0.036*cos(45) + 0.144 = 0.17 N
0.144*sin(45) - 0.072 = -0.05 N
0.17 + 0.05 = F = 0.18 N

8. Fg = 9.8 * (1*10^-6)/ (tan(60) = (9*10^9)* (q^2)/(0.0625)
q= 6.3 *10 ^-9

Ian Ruiz

2017-05-05 16:13:35 UTC

6) The upmost or initial central charge holds the largest net force acting upon it. The other points produce no net force since their positive charges repel one another and fail to act on center point.
7) k*(2q^2)(a^2)=.072N
k*(2*q^2)/(sqrt(2a))^2=.036N
k*(4*q^2)/(a^2)=.144N
.036*cos(45)+.144=.17N
.144*sin(45)-.072=-.05N
.17N+.05N=.18N
8) Fg=9.8(1 x 10^-6)/tan(60)=(9 x 10^9) * (q^2)/(.0625)-->
q=6.3 x 10^-9