A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

Options

A. `400%`
B. `66.6%`
C. `33.3%`
D. `200%`

Answer

B

Explanation

`C=(epsilon_(0)A)/(d)` (original)
`C_(1)=(K epsilon_(0)A)/(d//2)=(epsilon_(0)A)/(d//2K)=(epsilon_(0)A)/(d//10), C_(2)=(epsilon_(0)A)/(d//2)`

`therefore (1)/(C')=(1)/(C_(1))+(1)/(C_(2))=(d//10)/(epsilon_(0)A)+(d//2)/(epsilon_(0)A)=((3d//5))/(epsilon_(0)A)`
`rArr C''=(5)/(3)(epsilon_(0)A)/(4)`
`therefore` Increase in capacitance `=((5)/(3)(epsilon_(0)A)/(d)-(epsilon_(0)A)/(d))//(epsilon_(0)A)/(d)`
`=(5)/(3)-1=(2)/(3)i.e. 66.6%`.