Figure given below shows two identical parallel plate capacitors connected to a battery with switch

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2. Electric Potential and Capacitance

hard

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

$3:1$

$5:1$

$3:5$

$5:3$

(IIT-1983)

Solution

(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, $\frac{{{U_1}}}{{{U_2}}} = \frac{3}{5}$

Standard 12

Physics

A capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with dielectric, a charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

medium

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

hard

A parallel plate capacitor with plate separation $5$ $\mathrm{mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is_____.

hard

(JEE MAIN-2024)

Match the pairs

Capacitor Capacitance

$(A)$ Cylindrical capacitor $(i)$ ${4\pi { \in _0}R}$

$(B)$ Spherical capacitor $(ii)$ $\frac{{KA{ \in _0}}}{d}$

$(C)$ Parallel plate capacitor having dielectric between its plates $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$

$(D)$ Isolated spherical conductor $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} – {r_1}}}$

medium

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

medium

Capacitor	Capacitance
$(A)$ Cylindrical capacitor	$(i)$ ${4\pi { \in _0}R}$
$(B)$ Spherical capacitor	$(ii)$ $\frac{{KA{ \in _0}}}{d}$
$(C)$ Parallel plate capacitor having dielectric between its plates	$(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$
$(D)$ Isolated spherical conductor	$(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} – {r_1}}}$

Solution

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