Question

A parallel plate capacitor of capacity $C_0$ is charged by a battery. After charging, battery is disconnected and a dielectric is filled between the plates of capacitor. Column-I represents quantity and column-II represents the change occurred.

Column I		Column II
i.	Potential energy of capacitor	p.	increases
ii.	Potential difference between plates	q.	decreases
iii.	Capacity of capacitor	r.	remains same
iv.	Charge on capacitor	s.	may increase or decrease

Match the columns and choose correct option from the given codes.

• A. i-(s), ii-(r), iii-(q), iv-(p)
• B. i - (q), ii - (q), iii-(p), iv-(r)
• C. i-(p), ii-(q), iii-(r), iv-(s)
• D. i-(r), ii-(s), iii-(p), iv-(q)

Answer 1

Answer: (B)

As battery is disconnected, $Q$ remains same.
$C^{\prime }=\frac{{\epsilon }_{0}KA}{d}=KC\Rightarrow C$ increases
Now, $V^{\prime }=\frac{Q^{\prime }}{C^{\prime }}=\frac{Q}{KC}=\frac{V}{K}$
$\therefore$ Potential difference decreases
Now, $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^{{}^{\prime }2}=\frac{1}{2}KC\times \frac{V^2}{K^2}=\frac{U}{K}$
$\therefore$ Potential energy is also reduced