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  • Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch 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

Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch 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.

 

 
 
 
 
 

Answers (1)

Net capacitance when switch S is closed

=2C

Energy stored =\frac{1}{2}\times (2C)V^{2}=CV^{2}

When the switch is opened,

Voltage of B

V^{1}=\frac{Q}{KC}

Energy stored

=\frac{1}{2}KCV^{2}+\frac{1}{2}\frac{Q^{2}}{KC}

=\frac{1}{2}KCV^{2}+\frac{1}{2}\frac{CV^{2}}{K}

=\frac{1}{2}CV^{2} (\frac{K^{2}+1}{K})

The required ratio

=\frac{2k}{k^{2}+1}

Posted by

Safeer PP

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