11.3 Solutions
517
11.67 For an infinite non-conducting charge sheet E = σ/2ε 0
∴
σ = 2Eε 0 = 2 × 200 × 8.85 × 10
−12
= 3.54 × 10
−9 C/m
2
The electric field is independent of the distance.
For an infinite conducting sheet
σ = Eε 0 = 200 × 8.85 × 10
−12
= 1.77 × 10
−9 C/m
2
11.3.3 Capacitors
11.68 The field strength E 0 between the plates of a parallel plate capacitor in vacuum is
E 0 =
V
d
=
σ
ε 0
(1)
where V is the applied voltage, σ the charge density (charge per unit area)
and ε 0 the permittivity in vacuum. Now σ = q/A. Therefore the capacitance
in air or vacuum will be
C 0 =
q
V
=
Aε 0
d
(2)
With the introduction of the dielectric slab the electric field in the slab will
be E 0 /K , and the potential across the capacitor becomes
V = E 0 (d − K ) +
E 0 t
K
= E 0
(d − t) +
t
K
=
q
Aε 0
(d − t) +
t
K
∴ C =
q
V
=
ε 0 A
d − t
1 −
1
K
(3)
If a metal of thickness t is to be introduced, the effective distance between
the capacitor plates is reduced and the capacitance becomes
C =
ε 0 A
d − t
(4)
a result which is obtained by putting K = ∞ in (3).