A parallel plate air capacitor has a capacitance C. The capacitor is half-filled with a dielectric of dielectric constant 5. The percentage increase in the capacitance must have been -- %.

Answer

`66.67`

Explanation

`C=(epsilon_0A)/d`
When it is half-filled with diclectric of `K=5` , its capacitance becomes `C_s`.

Now,
`C_1 =(Kepsilon_0A)/((d//2))=(epsilon_0A)/((d//2K))=(Kepsilon_0A)/((d//10))`
`C_2 =(epsilon_0A)/((d//2))`
`:. 1/C_s =1/C_1 + 1/C_2 =((d//10))/(epsilon_0A)+((d//2))/(epsilon_0A)=((3d//5))/(epsilon_0A)`
`implies C_s = 5/3 xx(epsilon_0A)/d =5/3C` Percentange increase in capacitance
`=(C_s-C)/C xx 100`
`=(5/3 C-C)/C xx 100`
`=66.67%`