Two Different Things

Electricity and Magnetism Level 1

Two identical parallel plate capacitors are connected to a battery with the switch S S closed. When S S is opened and the free space between the capacitors is filled with a material of dielectric constant K K , the ratio of the total electrostatic energies stored in both capacitors, after and before the introduction of the dielectric, is __________ \text{\_\_\_\_\_\_\_\_\_\_} .

1 4 ( K + 3 K ) \dfrac{1}{4}\left(K+\frac{3}{K}\right) 1 6 ( K + 3 2 K ) \dfrac{1}{6}\left(K+\frac{3}{2K}\right) 1 2 ( K + 6 K ) \dfrac{1}{2}\left(K+\frac{6}{K}\right) 1 4 ( K + 4 5 K ) \dfrac{1}{4}\left(K+\frac{4}{5K}\right)

Rohit Ner by Rohit Ner , 22, India

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1 solution

Parth Sankhe
Oct 12, 2018

Energy before dielectric = 1 2 ( C + 3 C ) V 2 \frac {1}{2} (C+3C)V^2

After S is closed, the charge on the upper capacitor will be conserved(as it is no longer connected to a battery), while the voltage on the lower one will be conserved(as it will still be connected to the battery).

Q 1 = 3 C V = c o n s t a n t Q_{1}=3CV=constant

(As the dielectric is introduced, the capacitance of each capacitor gets multiplied by it's directric constant)

New energy in the upper capacitor = Q 1 2 2 C 1 = 9 C 2 V 2 2 ( 3 K C ) = 3 C V 2 2 K \frac {Q_{1}^2}{2C_{1}}=\frac {9C^2V^2}{2(3KC)}=\frac {3CV^2}{2K}

New energy in the lower capacitor = 1 2 ( K C ) V 2 \frac {1}{2}(KC)V^{2}

Adding the new energies and dividing by the original energy, we have the answer.

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