Solution:

Ans. Equivalent capacitance in series

Energy in series combination $1 / 2$

Charge in series combination $1 / 2$

Equivalent capacitance in parallel combination $1 / 2$

Energy in parallel combination $1 / 2$ Charge in parallel combination $1 / 2$

In series combination :

$$ \frac{1}{C_{s}}=\left(\frac{1}{12}+\frac{1}{12}\right)(p F)^{-1} \quad 1 / 2 $$

$\therefore \quad C_{\mathrm{s}}=6 \times 10^{-12} \mathrm{~F}$

$U_{\mathrm{s}}=\frac{1}{2} \mathrm{CV}^{2}$

$U_{\mathrm{s}}=\frac{1}{2} \times 6 \times 10^{-12} \times 50 \times 50 \mathrm{~J}$

$$ \begin{aligned} U_{\mathrm{s}} & =75 \times 10^{-10} \mathrm{~J} \ q_{\mathrm{s}} & =C_{\mathrm{s}} V \end{aligned} $$

$$ =6 \times 10^{-12} \times 50 \mathrm{C} $$

$$ \begin{aligned} & =300 \times 10^{-12} \mathrm{C} \ & =3 \times 10^{-10} \mathrm{C} \end{aligned} $$

$1 / 2$

In parallel combination :

$$ \begin{aligned} C_{p} & =(12+12) p \mathrm{~F} \ C_{p} & =24 \times 10^{-12} \mathrm{~F} \ U_{p} & =\frac{1}{2} \times 24 \times 10^{-12} \times 2500 \mathrm{~J} \ U_{p} & =3 \times 10^{-8} \mathrm{~J} \ q_{p} & =C_{p} V \ & =24 \times 10^{-12} \times 50 \mathrm{C} \ & =1.2 \times 10^{-9} \mathrm{C} \end{aligned} $$

$$ \therefore \quad C_{p}=24 \times 10^{-12} \mathrm{~F} $$

[CBSE Marking Scheme 2017]

13. Two identical parallel plate capacitors $A$ and $B$ are connected to a battery of $V$ volts with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $\kappa$. Find the ratio of the total electrostatic energy stored in bothcapacitors before and after the introduction of the dielectric.

[Topper’s Answers 2017]

[AI Q. 14. Two parallel plate capacitors $X$ and $Y$ have the same area of plates and same separation between them. $X$ has air between the plates while $Y$ contains a dielectric medium of $\varepsilon_{r}=4$.

(i) Calculate capacitance of each capaciton i equivalent capacitance of the combination is $4 \mu \mathrm{F}$.

(ii) Calculate the potential difference between the plates of $X$ and $Y$.

(iii) Estimate the ratio of electrostaticengy stored in $X$ and $Y$.

U] [Delhi I, II, III 2016, Set D, 2009, Set F 2008]

Ans. (i) Let

$$ \begin{aligned} & C_{X}=C \ & C_{Y}=4 C \end{aligned} $$

(as it has a dielectricmedium of $\varepsilon_{r}=4$ )

For series combination of two capacitors

$$ \frac{1}{C}=\frac{1}{C_{X}}+\frac{1}{C_{Y}} $$

$\Rightarrow \quad \frac{1}{4 \mu \mathrm{F}}=\frac{1}{\mathrm{C}}+\frac{1}{4 \mathrm{C}}$

$$ \frac{1}{4 \mu \mathrm{F}}=\frac{5}{4 \mathrm{C}} $$

$\Rightarrow \quad C=5 \mu \mathrm{F}$

Hence $\quad C_{X}=5 \mu \mathrm{F}$

(ii) Total charge

$$ C_{Y}=20 \mu \mathrm{F} $$

$$ \begin{aligned} & =4 \mu \mathrm{F} \times 15 \mathrm{~V}=60 \mu \mathrm{C} \ V_{X} & =\frac{Q}{C_{X}}=\frac{60 \mu \mathrm{C}}{5 \mu \mathrm{F}}=12 \mathrm{~V} \end{aligned} $$

$$ V_{Y}=\frac{Q}{C_{Y}}=\frac{60 \mu \mathrm{C}}{20 \mu \mathrm{F}}=3 \mathrm{~V} $$

(iii)

$$ =\frac{\overline{2 C_{X}}}{\frac{Q^{2}}{2 C_{x}}}=\frac{C_{Y}}{C_{X}}=\frac{20}{5}=4: 1 \quad \mathbf{1} $$

(Also accept any other correct alternative method)

[CBSE Marking Scheme 2016]

. A parallel plate capacitor, of capacitance $20 \mu \mathrm{F}$, is connected to a $100 \mathrm{~V}$, supply. After sometime, the battery is disconnected, and the space, between the plates of the capacitor is filled with a dielectric of dielectric constant 5 . Calculate the energy stored in the capacitor.

(i) Before

(ii) After the dielectric has been put in between its plates.

A [O.D. Comptt. I, II, III, 2016]

Ans. Charge stored, $Q=C V=20 \times 10^{-6} \times 100 \mathrm{C} \quad 1 / 2$

$$ =2000 \mu \mathrm{C} $$

New value of capacitance

$$ \begin{aligned} & =5 \times 20 \mu \mathrm{F} \ & =100 \mu \mathrm{F} \end{aligned} $$

Energy stored in a capacitor

$$ =\frac{1}{2} \frac{Q^{2}}{C}\left(=\frac{1}{2} C V^{2}=\frac{1}{2} Q V\right) \quad 1 / 2 $$

(i) Energy stored before dielectric

$$ \begin{align*} & =\frac{1}{2} \times \frac{\left[2000 \times 10^{-6}\right] \times\left(2000 \times 10^{-6}\right)}{20 \times 10^{-6}} \ & =0.1 \mathrm{~J} \tag{1} \end{align*} $$

(ii) Energy stored after the dielectric is introduced ( $\because$ there is no change in the value of $Q$ )

$$ \begin{align*} & =\frac{1}{2} \times \frac{2000 \times 10^{-6} \times 2000 \times 10^{-6}}{100 \times 10^{-6}} 1 / 2 \ & =0.02 \mathrm{~J} \ & 1 / 2 \end{align*} $$

[CBSE Marking Scheme, 2016]