1. Examine the charge distribution in the diagram below. (picutre attached) a) Determine the net force charge acting at q1 (+ 2.0 × 10-5C), caused by q2 (-4.0 × 10-5C) and q3 (-4.0 × 10-5C). Be sure to include a free body diagram representing the force acting at q1 in your solution. b) Determine the net electric field acting at q1.
1. Examine the charge distribution in the diagram below. (picutre attached) a) Determine the net force charge acting at q1 (+ 2.0 × 10-5C), caused by q2 (-4.0 × 10-5C) and q3 (-4.0 × 10-5C). Be sure to include a free body diagram representing the force acting at q1 in your solution. b) Determine the net electric field acting at q1.
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1. Examine the charge distribution in the diagram below. (picutre attached)
a) Determine the net force charge acting at q1 (+ 2.0 × 10-5C), caused by q2 (-4.0 × 10-5C) and q3 (-4.0 × 10-5C). Be sure to include a free body diagram representing the force acting at q1 in your solution.
b) Determine the net electric field acting at q1.
I attached my answers but I did not know how to draw the free body diagram. I also didn't know if the net electric field was a vector quantity. If you could check this as well as show me how to do the free body diagram.
![3. a) q2 =- 4.0 × 10 °C
93 =- 4.0 × 10 c
b) q, =-4.0 x 10 °C
93 =- 4.0 × 10°C
r2 = 24cm = 0.24m
ra = 24cm = 0.24m
r, = 24cm = 0.24m
r = 24cm = 0.24m
k = 9.0 × 10 °N -
Enet = ?
k = 9.0 × 10 °N "L
Enet = ?
(9.0×10°NX-4.0x10 *c)
(0.24m)
=- 6.25 x 10 “N/C
legl =
E/ = 4
(9.0x10 °NX-4.0×10°C)
(9.0×10 °N)-4.0×10 °C)
(0.24m) ?
6.25 × 10 “N/C
(0.24m)
=- 6.25 x 10 “N/C
kq3
|ɛ3| =
Ener = V(- 6.25 × 10 “N/C) 2 + (- 6.25 × 10 “N/C) ?
= V7.82 x 10 °N ²/C?
(9.0x10 'N뜰(-4.0x 10 -C)
(0.24m) ²
=- 6.25 × 10 6N/C
= 8.84 x 10 °NIC
q = 2.0 × 10 C
E = 8.84 x 10 “N/C
F = ?
Enet
(-6.25 × 10 °N/C) ² + (- 6.25 × 10 “N/C)?
V7.82 × 10 ³N ²/c ²
FE = që
= 8.84 × 10 °N/C
= (2.0 x 10 °C )(8.84 x 10 “N/C)
= 1.8 x 10?N
tane = -6.25×10 “N/C
-6.25x10 NIC
tane = 625x10*N/C
-6.25x10 "N/C
0 = 90°
e = 90°
Therefore, the net electric charge acting at q, is 8.84 × 10 °N/C [E 90°N].
Therefore, the net force charge at q, is 1.8 × 10 ²N [E 90 ° N].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13f02598-3671-4d97-8c50-c9caed302e6d%2F980776eb-d385-4399-91be-df6a6e9c2324%2Fz7t3z2_processed.png&w=3840&q=75)
Transcribed Image Text:3. a) q2 =- 4.0 × 10 °C
93 =- 4.0 × 10 c
b) q, =-4.0 x 10 °C
93 =- 4.0 × 10°C
r2 = 24cm = 0.24m
ra = 24cm = 0.24m
r, = 24cm = 0.24m
r = 24cm = 0.24m
k = 9.0 × 10 °N -
Enet = ?
k = 9.0 × 10 °N "L
Enet = ?
(9.0×10°NX-4.0x10 *c)
(0.24m)
=- 6.25 x 10 “N/C
legl =
E/ = 4
(9.0x10 °NX-4.0×10°C)
(9.0×10 °N)-4.0×10 °C)
(0.24m) ?
6.25 × 10 “N/C
(0.24m)
=- 6.25 x 10 “N/C
kq3
|ɛ3| =
Ener = V(- 6.25 × 10 “N/C) 2 + (- 6.25 × 10 “N/C) ?
= V7.82 x 10 °N ²/C?
(9.0x10 'N뜰(-4.0x 10 -C)
(0.24m) ²
=- 6.25 × 10 6N/C
= 8.84 x 10 °NIC
q = 2.0 × 10 C
E = 8.84 x 10 “N/C
F = ?
Enet
(-6.25 × 10 °N/C) ² + (- 6.25 × 10 “N/C)?
V7.82 × 10 ³N ²/c ²
FE = që
= 8.84 × 10 °N/C
= (2.0 x 10 °C )(8.84 x 10 “N/C)
= 1.8 x 10?N
tane = -6.25×10 “N/C
-6.25x10 NIC
tane = 625x10*N/C
-6.25x10 "N/C
0 = 90°
e = 90°
Therefore, the net electric charge acting at q, is 8.84 × 10 °N/C [E 90°N].
Therefore, the net force charge at q, is 1.8 × 10 ²N [E 90 ° N].
Transcribed Image Text:- 4.0 x 10-5 C
N
93
24cm
q1
+
q2
24cm
2
+2.0 x 105 C
- 4.0 x 105 C
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