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. 4
B. `66.6%`
C. `33.3%`
D. 2

Answer

B

Explanation

`C = (epsilon_(0)A)/(d)` …(i)
When it is half filled with dielectric of K = 5, its capacitance becomes `C_(s)` which is series combination of
`C_(1) = (K epsilon_(0)A)/(d//2) = (epsilon_(0)A)/(d//2K) = (epsilon_(0)A)/(d//10)`
and `C_(2) = (epsilon_(0)A)/(d//2)`
`:. (1)/(C_(S)) = (1)/(C_(1)) + (1)/(C_(2)) = (d//10)/(epsilon_(0)A) + (d//2)/(epsilon_(0)A) = ((3d//5))/(epsilon_(0)A)`
`rArr C_(s) = (5)/(3)(epsilon_(0)A)/(d) = (5C)/(3)` (Using (i))
Percentange increase in capacitance
`= (C_(s) - C)/(C ) xx 100 = ((5)/(3)C-C)/(C ) xx 100 = (2)/(3) xx 100 = 66.6%`