Two identical capacitors 1 and 2 are connecte | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the capacitance of each identical capacitor be $C$. When the dielectric slab of constant $k$ is inserted in capacitor $2$,its capacitance beco

Vedclass · Accepted Answer

$\frac{Q'_2}{Q_2} = \frac{k + 1}{2k}$

Vedclass · Answer

(C) Let the capacitance of each identical capacitor be $C$. When the dielectric slab of constant $k$ is inserted in capacitor $2$,its capacitance becomes $kC$. Since the capacitors are in series,the equivalent capacitance is $C_{eq} = \frac{C \cdot kC}{C + kC} = \frac{kC}{k + 1}$. The charge on each capacitor in series is equal to the total charge supplied by the battery: $Q_1 = Q_2 = Q_{eq} = C_{eq}E = \frac{kCE}{k + 1}$. When the dielectric is removed,the capacitance of capacitor $2$ becomes $C$. The new equivalent capacitance is $C'_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$. The new charge on each capacitor is $Q'_1 = Q'_2 = Q'_{eq} = C'_{eq}E = \frac{CE}{2}$. Now,calculating the ratio: $\frac{Q'_1}{Q_1} = \frac{Q'_2}{Q_2} = \frac{CE/2}{kCE/(k+1)} = \frac{k+1}{2k}$. Comparing this with the given options,we see that option $C$ matches this result.

Similar Questions

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module