A parallel plate capacitor of capacity `5 mu F` and plate separation `6 cm` is connected to a `1V` battery and is charged. A dielectric of dielectric constant `4` and thickness `4 cm` is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is.

Options

A. `2 mu C`
B. `3 mu C`
C. `5 muC`
D. `10 mu C`

Answer

C

Explanation

Charge on capacitor plates without the dielectric is
`Q =CV = (5 xx 10^(-6)F) xx 1V = 5 xx 10^(-6)C = 5 mu C`
The capacitance after the dielectric is introduced is
`C. = (epsilon_(0)A)/(d-(t-(t)/(K))) = (epsilon_(0)A//d)/((1-(t-(t)/(K))/(d)))`
`= (c )/(1-((t-(t)/(K))/(d)))= (5mu F)/(1-((4cm-(4cm)/(4))/(6cm))) = (5mu F)/(1-((4-1)/(6))) = 10 mu F`
`:.` Charge on capacitor plates not will be
`Q. = C.V = 10 mu F xx 1V = 10 mu C`
Additional charge transferred `= Q. - Q = 10 mu C - 5 mu C = 5 mu C`