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].
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].
- 4.0 x 10-5 C
N
93
24cm
q1
+
q2
24cm
2
+2.0 x 105 C
- 4.0 x 105 C
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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