11.3 Solutions
485
tan θ =
F
mg
=
4 × 10 −4
7.84 × 10 −4 = 0.51
θ = 27
◦
Squaring (1) and (2) and adding and extracting the square root
T =
F 2 + (mg) 2 =
(4 × 10 −4 ) 2 + (7.84 × 10 −4 ) 2 = 8.8 × 10
−4 N
11.7 Electric potential
V =
1
4πε 0
q
1
+
q
2
+
q
4
+
q
8
+ · · ·
=
q
4πε 0
1 +
1
2
+
1
2 2 +
1
2 3 + · · ·
The terms in brackets form a geometric progression of infinite terms whose
sum is
S =
a
1 − r
=
1
1 −
1
2
= 2
∴ V =
q
2πε 0
Electric field
E =
1
4πε 0
q
1
+
q
2 2 +
q
4 2 +
q
8 2 + · · ·
=
1
4πε 0
q
1 −
1
4
=
q
3πε 0
11.8 V =
1
4πε 0
q −
q
2
+
q
4
−
q
8
+
q
16
−
q
32
+ · · ·
=
q
4πε 0
1 +
1
4
+
1
16
+ · · ·
−
q
8πε 0
1 +
1
4
+
1
16
+ · · ·
=
1
4πε 0
q −
q
2
1 +
1
4
+
1
16
+ · · ·
=
q
8πε 0
1
1 −
1
4
=
q
6πε 0