Spark image

The action of a dielectric

When a dielectric material is placed between the plates of a parallel-plate (or other) capacitor the capacitance increases. The reason for this is shown in Figures 1(a) and 1(b).




The charges on the plates of the capacitor induce opposite charges on the two surfaces of the dielectric. This has the effect of reducing the potential difference across the capacitor.


This can be explained as follows. Under the action of the electric field within the dielectric the molecules become polarised and are aligned as shown in Figure 1(b).
Since the capacitance of a capacitor is given by the formula C = Q/V, if the p.d. (V) across the capacitor is reduced the capacitance must be increased.

The capacitance of a parallel-plate capacitor with a material of relative permittivity er filling the space between the plates is

Capacitance (C) = eoerA/d


where er is the ratio of the capacitances when the space between the plates is a vacuum or a dielectric.

The capacitance of a capacitor when only a thickness t of the air space is filled is

C = eoerA/[er(d - t) + t]


The following example should illustrate how to deal with some of the problems of dielectrics and capacitors.


Example problems
A parallel-plate capacitor has an area of 100 cm2, a plate separation of 1.0 cm and is charged initially to a potential of 100 V (call this V0). The supply is disconnected and a slab of dielectric, 0.5 cm thick and of relative permittivity 7, is then placed between the plates as shown in Figure 2.
Calculate
(a) the capacitance C0 before the slab is inserted,
(b) the charge on the plates Q
(c) the electric field strength in the gap between the plates and the dielectric
(d) the electric field strength within the dielectric
(e) the potential difference between the plates after the dielectric is inserted
(f) the capacitance when the dielectric has been inserted.

(a) C = eoA/d = [8.9 x 10-12 x 10-2]/10-2 = 8.9 x 10-12 F.
(b) Q = CoV0 = [8.9x10-12 x 100] = 8.9 x 10-10C.
(c) Eo = Q/eoA = 8.9 x 10-10/[8.9 x 10-12 x10-2] = 104 Vm-1
(d) E1 = Q/eA = Eo/er = 104/7 = 0.14 x104 Vm-1
(e) p.d. = [0.5 x 104 + 0.5 x 0.14 x 104]/100 = 57 V
(f) C = eA/[er(d t) + t] = 7eoA/0.04 = 15.2 x 10-12 F.

 
 
 
© Keith Gibbs 2011