An air filled parallel plate capacitor initially has a capacitance of C- 2.07 nF

Question

An air filled parallel plate capacitor initially has a capacitance of C- 2.07 nF. Then the air gap between the plates is filled with two different dielectric materials with equal volume but with relative permittivities of ?1 = 2.95 and ?2-9.98 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 K Submit Answer Tries 0/99 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? Submit Answer Tries 0/99

Explanation / Answer

Given,

C = 2.07 nF

e1 = 2.95 ; e2 = 9.98

Now as seen in the figure, there are now two capacitors with dielectrics in series withgap halved

C1 = e1 C = 2.95 x 2.07 x 2 = 12.21 nF

C2 = e2 C = 9.98 x 2.07 x 2 = 41.32 nF

C = C1C2/(C1 + C2)

C = 12.21 x 41.32/(12.21 + 41.32) = 9.42 nF

Hence, C = 9.42 nF

In the later case, C1 and C2 are in parallel and area halved

C1 = 2.95 x 2.07/2 = 3.05 nF

C2 = 9.98 x 2.07/2 = 10.33

C = C1 + C2

C = 3.05+10.33 = 13.38 nF

Hence, C = 13.38 nF

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