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STD 12 - 2. Electric potential and capacitance — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electric potential and capacitance4 Marks
MCQ
Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
  • A
    $3:1$
  • B
    $5:1$
  • ✓
    $3:5$
  • D
    $5:3$
✓

Answer

Correct option: C.
$3:5$
c
(c) Initially potential difference across both the capacitor is same hence energy of the system is
${U_1} = \frac{1}{2}C{V^2} + \frac{1}{2}C{V^2} = C{V^2}$$……(i)$
In the second case when key $K$ is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes $3C$, while potential difference across $A$ is $V$ and potential difference across $B$ is $\frac{V}{3}$ hence energy of the system now is
${U_2} = \frac{1}{2}\,(3C){V^2} + \frac{1}{2}\,(3C)\,{\left( {\frac{V}{3}} \right)^2}$$ = \frac{{10}}{6}\,C{V^2}$$……(ii)$
So, $\frac{{{U_1}}}{{{U_2}}} = \frac{3}{5}$

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