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Question: A parallel plate capacitor of capacitance \(5 \mu \mathrm {~F}\)and plate separation 6 cm is connect...

A parallel plate capacitor of capacitance 5μ F5 \mu \mathrm {~F}and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is.

A

2μC2 \mu \mathrm { C }

B

3μC3 \mu \mathrm { C }

C

5μC5 \mu \mathrm { C }

D

Answer

5μC5 \mu \mathrm { C }

Explanation

Solution

: Charge on capacitor plates without the dielectric is

The capacitance after the dielectric is introduced is

=5μ F1−(4−16)=10μ F= \frac { 5 \mu \mathrm {~F} } { 1 - \left( \frac { 4 - 1 } { 6 } \right) } = 10 \mu \mathrm {~F}

∴\therefore Charge on capacitor plates now will be,

Additional charge transferred =Q′−Q=10μC−5μC= \mathrm { Q } ^ { \prime } - \mathrm { Q } = 10 \mu \mathrm { C } - 5 \mu \mathrm { C }

=5μC= 5 \mu \mathrm { C }