⇒Electrostatics

Correct Option is (D)

Let the initial capacitance be ${\mathrm{C}}_0=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}$

Let the charge on the capacitor be $Q_{\text{initial }}=C_0{V}_0$

Plate separation is increased by 3 times i.e., ${\mathrm{d}}^{\prime }=3\mathrm{d}$

${\mathrm{C}}_{\text{final }}=\frac{{\epsilon }_0\mathrm{A}}{3\mathrm{d}}=\frac{1}{3}\left(\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}\right)=\frac{{\mathrm{C}}_0}{3}$

Let ${\mathrm{Q}}_{\text{final }}$ be the final charge on the capacitor and $V_{\text{final }}$ be the final potential on the capacitor.

$\therefore {\mathrm{Q}}_{\text{final }}={\mathrm{C}}_{\text{final }}{\mathrm{V}}_{\text{final }}=\frac{1}{3}{\mathrm{C}}_0{\mathrm{V}}_{\text{final }}$

As the capacitor is isolated,

$\begin{array}{rl}& {\mathrm{Q}}_{\text{final }}={\mathrm{Q}}_{\text{initial }}\text{, }\\ & {\mathrm{C}}_0{\mathrm{V}}_0=\frac{1}{3}{\mathrm{C}}_0{\mathrm{V}}_{\text{final }}\\ & \therefore {\mathrm{V}}_{\text{final }}=3{\mathrm{V}}_0\\ & \text{ Work done }=\text{ Final P.E }-\text{ Initial P.E }\\ & =\frac{1}{2}{\mathrm{C}}_{\text{final }}{\left({\mathrm{V}}_{\text{final }}\right)}^2-\frac{1}{2}{\mathrm{C}}_0{\left({\mathrm{V}}_0\right)}^2\\ & =\frac{1}{2}\frac{{\mathrm{C}}_0}{3}9{\mathrm{V}}_0^2-\frac{1}{2}{\mathrm{C}}_0{\mathrm{V}}_0^2\\ & ={\mathrm{C}}_0{\mathrm{V}}_0^2(\frac{3}{2}-\frac{1}{2})\\ & ={\mathrm{C}}_0{\mathrm{V}}_0^2\\ & =\frac{{\epsilon }_0{\mathrm{AV}}_0^2}{\mathrm{d}}\dots (\because {\mathrm{C}}_0=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}})\end{array}$

Correct Option is (B)

$\begin{array}{rl}& {\mathrm{C}}_1=3\mu \mathrm{F}\text{ and }{\mathrm{C}}_2=2\mu \mathrm{F}\\ \therefore & {\mathrm{C}}_{\text{series }}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{6}{5}\mu \mathrm{F}\\ & \text{ Also, }\mathrm{Q}=\mathrm{CV}\\ \mathrm{Q}& ={\mathrm{C}}_{\text{series }}\times \mathrm{V}\\ & =\frac{6}{5}\times 100=120\mu \mathrm{C}\end{array}$

Q will be the same across both the capacitors as they are in series.

$\therefore$ Potential across capacitors,

$\begin{array}{rl}& {\mathrm{V}}_1=\frac{\mathrm{Q}}{{\mathrm{C}}_1}=\frac{120}{3}=40\mathrm{V}\\ & {\mathrm{V}}_2=\frac{\mathrm{Q}}{{\mathrm{C}}_2}=\frac{120}{2}=60\mathrm{V}\\ \therefore {\mathrm{V}}_2& :{\mathrm{V}}_1=60:40=3:2\end{array}$

Correct Option is (A)

For a parallel plate capacitor, the energy density $=\frac{1}{2}{E}^2{\epsilon }_0$

But $\mathrm{E}=\frac{\sigma }{{\epsilon }_0}$

$\begin{array}{rl}\therefore \text{ Energy density }& =\frac{1}{2}\frac{{\sigma }^2}{{\epsilon }_0^2}\times {\epsilon }_0\\ & =\frac{{\sigma }^2}{2{\epsilon }_0}\\ & =\frac{{\left(\frac{\mathrm{q}}{\mathrm{A}}\right)}^2}{2{\epsilon }_0}\dots .(\because \sigma =\mathrm{q}/\mathrm{A})\\ & =\frac{{\mathrm{q}}^2}{2{\mathrm{A}}^2\cdot {\epsilon }_0}\end{array}$

Correct Option is (C)

We know, $\mathrm{Q}=\mathrm{C}.\mathrm{V}$

$\begin{array}{ll}\therefore Q_1=10\mathrm{C}\text{ and }{\mathrm{Q}}_2=5\mathrm{C}& \\ \text{ Energy stored, }\mathrm{E}=\frac{1}{2}{\mathrm{CV}}^2\\ \therefore {\mathrm{E}}_1=\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}^2=\frac{1}{2}\times 2\mathrm{C}\times 25=25\mathrm{J}\\ \text{ Similarly, }{\mathrm{E}}_2=\frac{1}{2}{\mathrm{C}}_2{\mathrm{V}}^2=\frac{1}{2}\times \mathrm{C}\times 25=12.5\mathrm{J}\\ \therefore \frac{{\mathrm{E}}_1-{\mathrm{E}}_2}{{\mathrm{Q}}_1-{\mathrm{Q}}_2}=\frac{12.5}{5}=\frac{5}{2}\end{array}$

Correct Option is (B)

Given $C=\frac{{\epsilon }_{0}A}{d}=2.5\mu F$

When half filled with air,

${\mathrm{C}}_1=\frac{{\epsilon }_0(\mathrm{A}/2)}{\mathrm{d}}=\frac{{\epsilon }_0\mathrm{A}}{2\mathrm{d}}\dots ..(\because {\epsilon }_{\mathrm{r}}=1)$

When half filled with a dielectric,

${\mathrm{C}}_2=\frac{{\epsilon }_{\mathrm{r}}{\epsilon }_0(\mathrm{A}/2)}{\mathrm{d}}=\frac{{\epsilon }_{\mathrm{r}}{\epsilon }_0\mathrm{A}}{2\mathrm{d}}$

From the figure, it can be seen that $C_1$ and $C_2$ are in parallel configuration.

$\begin{array}{ll}\therefore & {\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2\\ & \text{ Given, }{\mathrm{C}}_{\mathrm{eq}}=5\mu \mathrm{F}\\ & \Rightarrow 5\mu \mathrm{F}=\frac{{\epsilon }_0\mathrm{A}}{2\mathrm{d}}+\frac{{\epsilon }_{\mathrm{r}}{\epsilon }_0\mathrm{A}}{2\mathrm{d}}\\ & 5=\frac{2.5}{2}+{\epsilon }_{\mathrm{r}}\frac{2.5}{2}\\ & \frac{3.75}{1.25}={\epsilon }_{\mathrm{r}}\\ \therefore & {\epsilon }_{\mathrm{r}}=3\end{array}$