In the second case the dielectric materials are formed into two blocks next two
ID: 2274924 • Letter: I
Question
In the second case the dielectric materials are formed into two blocks next two each other. (See figure.) What is the capacitance in this case?
An air filled parallel plate capacitor initially has a capacitance of C = 5.58 nF. Then the air gap between the plates is filled with two different dielectric materials with equal volume but with relative permittivities of ?1 = 2.14 and ?2 = 9.64 respectively. In the first case the dielectric materials are formed into two layers parallel to the plates of the capacitor. (See figure.) Determine the capacitance of the capacitor. (Note: the relative permittivity denoted by? Is equivalent to the dielectric constant denoted by? In the second case the dielectric materials are formed into two blocks next two each other.(See figure.) What is the capacitance in this case?Explanation / Answer
This becomes two capacitors in parallel.
Capacitance of each such capacitor assuming it to be air-filled=2C=11.16nF
Capacitances are therefore, 11.16*2.14 and11.16*9.64= 23.88 nF and 107.58 nF
The equivalent capacitance= 23.88*107.58/(23.88+107.58)=19.58nF