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A parallel plate capacitor of capacity 5...

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.

A

`2 muC`

B

`3 muC`

C

`5 mu C`

D

`10 muC`

Text Solution

Verified by Experts

The correct Answer is:
C
Charge on capacitor plantes without the diectric is
`Q = CV = (5 xx 10^(-6)F) xx 1V = 5 xx 10^(-6) C = 5 mu`
The capacitance after the dielectric is introduced is
`C = (epsi_(0)A)/(d-(t-(t)/(k))) = (epsi_(0)A//d)/(1-((t-(t)/(k))/(d)))`
` = (C)/(1-((t-(t)/(k))/(d))) = (5 muF)/(1-((4cm -(4cm)/(4))/(6cm)))`
` = (5muF)/(1-((4-1)/(6))) = 10 muF`
`therefore` charge on capacitor plates now will be
`Q = C'V = 10 muF xx 1 V = 10 muC`
Additional charge transfered ` = Q - Q = 10 muC - 5 muC`
` = 5 mu C`
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