CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

Question : 22 of 26

Marks: +1, -0

Two identical parallel plate capacitors

{A}

and

B

are connected to a battery of

V

volts with the switch

\bo {S}

closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant

K

. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Solution:

Formula for energy stored
Energy stored before
Energy stored after
Ratio
Energy stored

=\;{1}/{2} C V^2 (=\;{1}/{2} \;{Q^2}/{C})

Net capacitance with switch

{\S}

closed

\; \;\;\;\; = {\C}+ {\C}=2 {\C}

Energy stored

=\;{1}/{2} × 2 {\C} × V^2=C V^2

After the switch S is opened, capacitance each capacitor of

=K C

∴

Energy stored in capacitor

{\A}=\;{1}/{2} K C V^2

For capacitor B,
Energy stored

=\;{1}/{2} \;{Q^2}/{K C}=\;{1}/{2} \;{C^2 V^2}/{K C}

=\;{1}/{2} \;{C V^2}/{K}

∴

Total Energy stored

\;=\;{1}/{2} K C V^2+\;{1}/{2} \;{C V^2}/{K}

=\;{1}/{2} C V^2 (K+\;{1}/{K})

\;=\;{1}/{2} C V^2 (\;{K^2+1}/{K})

∴

Required ratio

=\;{2 C V^2 ⋅ K}/{C V^2 (K^2+1) }=\;{2 K}/{ (K^2+1) }

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