electrostatic-potential-and-capacitance Question 31
Question: Q. 6. The given graph shows variation of charge $q$ versus potential difference $V$ for two capacitors $C_{1}$ and $C_{2}$. Both the capacitors have same plate separation but plate area of $C_{2}$ is greater than that of $C_{1}$. Which line ( $A$ or $B$ ) corresponds to $C_{1}$ and why?
[Comptt. O.D. I, II, III 2014]
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Solution:
Ans. For $C_{1}$
Line $B$
1
Since slope $(q / V)$ of Line $B$ is lesser than that of Line $A$.
[CBSE Marking Scheme 2014] 1 (AI Q. 1. 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 both capacitors before and after the introduction of the dielectric.
U [O.D. I, II, III 2017]
Ans. Formula for energy stored Energy stored before $1 / 2$
Energy stored after
Ratio
$$ \text { Energy stored }=\frac{1}{2} C V^{2}=\left(\frac{1}{2} \frac{Q^{2}}{C}\right) \quad 1 / 2 $$
Net capacitance with switch $S$ closed
$$ \begin{align*} & =C+C=2 C \ \text { Energy stored } & =\frac{1}{2} \times 2 C \times V^{2}=C V^{2} \end{align*} $$
After the switch $S$ is opened, capacitance of each capacitor $=\kappa \times C$
$\therefore$ Energy stored in capacitor $A$
$$ A=\frac{1}{2} \kappa C V^{2} $$
For capacitor $B$,
$$ \begin{aligned} \text { Energy stored } & =\frac{1}{2} \frac{Q^{2}}{\kappa C}=\frac{1}{2} \frac{C^{2} V^{2}}{\kappa C} \ & =\frac{1}{2} \frac{C V^{2}}{\kappa} \ \therefore \quad \text { Total Energy stored } & =\frac{1}{2} \kappa C V^{2}+\frac{C V^{2}}{\kappa} \ & =\frac{1}{2} C V^{2}\left(\frac{\kappa^{2}+1}{\kappa}\right) \ \text { Required ratio } & =\frac{2 C V^{2} \kappa}{C V^{2}\left(\kappa^{2}+1\right)} \ & =\frac{2 \kappa}{\left(\kappa^{2}+1\right)} \end{aligned} $$
[CBSE Marking Scheme 2017]
