Question

1. A parallel plate capacitor has an area of ​​1,000 cm ^ 2 and a distance between the plates of 1 cm. A dielectric material is inserted and its new capacitance is calculated to be 1.77 x 10 ^ -10 F. Determine the value of the dielectric constant.

a.	2
	b.	2.25
	c.	1.5
	d.	0.75

Answer 1

2 Area of plates of capacitor = A=1000 cm 2 ( 1 cm = 10-2 m =) (1 cm)2 = (1.-2m) 2 1 cm ² 10-4 m²) 2 = 2 4 A = 1000 x 6 m A = 0.1m? 2 - = Distance between plates Let dielectric constant of d = 1 cm lo in dielectric material =k →A Parallel plate capa cu for d & electric bermittivity of free space ε =8.854 Flm and 8-854 X 16-12 =C Capacitance with dielectric material c=1677 X lolof - Now, Capacitance with dielectric material is KE C = 2 d S cd 3 , - 2 -12 K = EA Now, using values in ③ we get - 10 1.77 x 10 F x lo K = 8-854 x 10 Fm xo.1 m² -12 xlo =) K =) K=1.999 z 2 0.8854 X 10-12 Fm So, dielectric Constant of material is 2 is option (a) Fm 1.77 = Hence, answer