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

Initial capacitance,

When it is half-filled by a dielectric of dielectric constant K, then
`C_(1)=(Kepsi_(0)A)/(d//2)=2K(epsi_(0)A)/(d)`
and `C_(2)=(epsi_(0)A)/(d//2)=(2epsi_(0)A)/(d)`
`:.(1)/(C')=(1)/(C_(1))+(1)/(C_(2))=(d)/(2epsi_(0)A)((1)/(K)+1)`
`=(d)/(2epsi_(0)A)((1)/(5)+1)=(6d)/(10epsi_(0)A)`
`:." "C'=(5epsi_(0)A)/(3d)`
Hence, increase in capacitance
`((5epsi_(0)A)/(3d)-(epsi_(0)A)/(d))/((epsi_(0)A)/(d))=(2)/(3)`
`=66.6%`