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Capacitors — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsCapacitors5 Marks
Question
Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.
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Answer


Initially when switch ‘s’ is closed
Total Initial Energy $=\Big(\frac{1}{2}\Big)\text{CV}^2+\Big(\frac{1}{2}\Big)\text{CV}^2=\text{CV}^2\ \dots(1)$
When switch is open the capacitance in each of capacitors varies, hence the energy also varies.
i.e. in case of ‘B’, the charge remains.
Same i.e. cv
$\text{C}_{\text{eff}}=3\text{C}$
$\text{E}=\frac{1}{2}\times\frac{\text{q}^2}{\text{c}}$
$=\frac{1}{2}\times\frac{\text{c}^2\text{v}^2}{3\text{c}}=\frac{\text{cv}^2}{6}$
In case of 'A'
$\text{C}_{\text{eff}}=3\text{c}$
$\text{E}=\frac{1}{2}\times\text{C}_{\text{eff}}\text{v}^2$
$=\frac{1}{2}\times\text{3c}\times\text{v}^2=\frac{3}{2}\text{cv}^2$
Total final energy $=\frac{\text{cv}^2}{6}+\frac{3\text{cv}^2}{2}=\frac{10\text{cv}^2}{6}$
Now, $\frac{\text{Initial Energy}}{\text{Final Energy}}=\frac{\text{cv}^2}{\frac{10\text{cv}^2}{6}}=3$

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