Answer:
a) 1.26e^-10F
b) 1.47e^-10F
c) 2.39e^-8C 2.89e^-8C
d) E=4500.94N/C
e) E'=5254.23N/C
f) 100.68V
g) 1.65e^-10J
Explanation:
To compute the capacitance we can use the formula:
where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.
(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:
(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:
(c)
The charge between the plates for both cases, with the slab is given by:
Q : without the slab
Q': with the slab
(d) The electric field between the plate is given by:
E: without the slab
E': with the slab
(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:
(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels: