1. F = F = Eq 0.08 = E (2 x 10-6) E = 40,000 N/C right q1 q21. -
F = Eq
0.08 = E (2 x 10-6) F
E = 40,000 N/C right 2. + -
q1 q2
(3 x 10-6)(6 x 10-6)
2.6 N attraction
F = k
= 9 x 109 = d 2
(0.25)2
This is the of the force on each charge
a) 2.6 N right
b) 2.6 N left

2. E1 = E3 = E2 = Electric Field (E) - vector Q1 + k d1 (4 x 10-6)
0.3 m Q1 3. +
E1 = k d1
.424 m 2
45°
0.3 m
(4 x 10-6)
E1= 9 x 109 - +
(0.3)2 q2 q3
E1 = 4 x 105 N/C right Q3 Q2
E3 = k
E2 = k d3 2 d2 2
(5 x 10-6)
(2 x 10-6)
E3= 9 x 109
E2= 9 x 109
(0.3)2
(0.424)2
E3 = 5 x 105 N/C up
E2 = 1.0 x ° below the negative x- axis

3. 5.4 x 105 N/C @ 52.3° above the +x-axis
Add the fields E1 E2
1.0 x 105
X = x 104
Y = x 104
4 x 105
45°
X = + 4 x 105
Y = 0 X Y
+4 x 105
-7.07 x 104
+5 x 105 E3
X = 0
Y = + 5 x 105
5 x 105
+3.29 x 105
+4.29 x 105
5.4 x ° above the +x-axis
4.29 x 105
3.29 x 105

4. 3000 more protons than electrons x = 1.6 x 10-19 C
V = V
d = 2.00 mm = m
m = 5.88 x kg 4.
Parallel Plates
V = E·d → = E (0.002)
E = 1.2 x 107 V/m up
Fg = mg → Fg = 5.88 x (9.8)
Fg= 5.76 x 10-9 N Fe +
Fe = Eq → 5.76 x 10-9 = 1.2 x 107 q Fg +
q = x C +
4.8 x C
1 proton
3000 more protons than electrons x =
1.6 x C
deficiency of 3000 electrons

5. V = 83,500 V m = 1.67 x 10-27 kg q = +1.6 x 10-19 C KEi = Wn5. +
Proton means
m = 1.67 x kg
q = +1.6 x C
The KE is converted to work done by the repulsive electrostatic force
PEi + KEi + Wc = PEf + KEf + Wn
KEi = Wn
½ mv2 = qV *
½ (1.67 x 10-27)(4 x 106)2 = (1.6 x 10-19)V
V = 83,500 V

6. Q
(1.6 x 10-19)
V = k
83,500 = 9 x 109 d
(d)
Divide by 9 x 109
(1.6 x 10-19)
9.28 x 10-6 =
(d)
Multiply by d
9.28 x 10-6 (d) = 1.6 x 10-19
d = 1.7 x m