11.3 Solutions
497
∴
dq
dx
=
3
2
2πε 0 mg
L
√
x
dq
dt
=
dq
dx
dx
dt
=
dq
dx
v =
dq
dx
a
√
x
=
3
2
a
2πε 0 mg
L
11.31 Consider an element of length dx of the thread at distance x from the
centre of the ring. The force between the ring and the element dx can be
resolved into x- and y-components, Fig. 11.32. The y-component will get
cancelled for reasons of symmetry. The field is entirely contributed by the
x-component. The charge in length dx is λdx. The electric force between the
wire and the ring is given by
F = F x =
1
4 πε 0
q λ dx cos θ
(R 2 + x 2 )
=
qλ
4πε 0
∞
0
xdx
(x 2 + R 2 ) 3/2
Put x = R cot θ , dx = −Rcosec 2 θ dθ
F = −
qλ
4πε 0 R
0
π/2
cos θ dθ =
qλ
4πε 0 R
Fig. 11.32
11.32 Consider an element of wire dx at distance x from O, Fig. 11.33. The charge
in dx will be λdx. The x-component of the electric field will be
E x =
E sin θ =
λdx sin θ
4πε 0 (x 2 + y 2 )
=
λ
4πε 0
∞
0
x dx
(x 2 + y 2 ) 3/2
Put x
2
+ y
2
= z
2
, x dx = z dz
where y = constant.
E x =
λ
4πε 0
∞
y
dz
z 2 =
λ
4πε 0 y
Similarly, E y =
E cos θ =
λ
4πε 0 y