A parallel plate capacitor has two plates of dimensions `10 cm xx 7 cm` separated by a distance of 0.7 mm. A glass plate of thickness 0.4 mm (dielectric constant = 6) and another dielectric medium of thickness 0.3 mm (dielectric constant = 2.5) are placed between the plates of the capacitor. Calculate the capacitance of the capacitor before and after introduction of the dielectric media.

Explanation

`A= 10cm xx 7cm = 70cm^2`
`A= 70 xx 10^(-4) m^2`
d=0.7mm

Case (i) Capacitance of the capacitor before
`C= (epsilon_0 A)/(d)`
`=(8.854xx10^(-12) xx 70 xx 10^(-4))/(0.7xx10^(-3))`
`=(619.78 xx 10^(-16))/(0.7 xx 10^(-3))`
`=885.4xx10^(-12)`
= 885.4 pF
Case (ii) Capacitance of the capacitor after introduction of dielectric media
`C.= (K_1K_2 epsilon_0 A)/(K_1d_2+K_2d_1)`
Here, `K_1 = 6, d_1 =0.4mm = 0.4xx10^(-3)m`
`K_2 = 2.5, d_2 = 0.3mm=0.3xx10^(-3)m`
`C.=(6xx2.5xx8.854xx10^(-12) xx 70xx10^(-4))/((6xx0.3xx10^(-3))+(2.5xx0.4xx10^(-3)))`
`C.=(9296.7xx10^(-16))/(1.8xx10^(-3) +1xc10^(-3))`
`=(9296.7xx10^(-16))/(2.8 xx 10^(-3))`
`= 3.320.25xx10^(-12)`
C. = 3320.25 pF