488
11 Electrostatics
tan θ =
F
mg
=
F
mg
1 −
ρ
ρ
F
F = K =
1
1 −
ρ
ρ
=
ρ
ρ − ρ =
1.6
1.6 − 0.8
= 2
Fig. 11.21
11.12 From the geometry of Fig. 11.22, AB = 0.1 m, OB = OC = 0.05
√
2 m,
AC = (0.1)
√
2 m.
Potential energy of q at B is U (B) =
q
4πε 0
q 1
AB
+
q 2
OB
Potential energy of q at C is U (C) =
q
4πε 0
q 1
AC
+
q 2
OC
Fig. 11.22
11 Electrostatics
tan θ =
F
mg
=
F
mg
1 −
ρ
ρ
F
F = K =
1
1 −
ρ
ρ
=
ρ
ρ − ρ =
1.6
1.6 − 0.8
= 2
Fig. 11.21
11.12 From the geometry of Fig. 11.22, AB = 0.1 m, OB = OC = 0.05
√
2 m,
AC = (0.1)
√
2 m.
Potential energy of q at B is U (B) =
q
4πε 0
q 1
AB
+
q 2
OB
Potential energy of q at C is U (C) =
q
4πε 0
q 1
AC
+
q 2
OC
Fig. 11.22