Two identical parallel plate capacitors of ca | vedclass

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

Question

(B) Initially,the equivalent capacitance is $C_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.The charge delivered by the battery is $Q_1 = C_{eq} E =

Vedclass · Accepted Answer

$\frac{k-1}{2(k+1)} \cdot C E$

Vedclass · Answer

(B) Initially,the equivalent capacitance is $C_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.The charge delivered by the battery is $Q_1 = C_{eq} E = \frac{C E}{2}$.When one of the capacitors is filled with a dielectric of constant $k$,its new capacitance becomes $kC$. The new equivalent capacitance of the series combination is:$C'_{eq} = \frac{C \cdot (kC)}{C + kC} = \left( \frac{k}{k+1} \right) C$.Since the battery remains connected,the new charge on the combination is:$Q_2 = C'_{eq} E = \left( \frac{k}{k+1} \right) C E$.The amount of charge that flows through the battery is the change in charge:$\Delta Q = Q_2 - Q_1 = \left( \frac{k}{k+1} - \frac{1}{2} \right) C E$.$\Delta Q = \left( \frac{2k - (k+1)}{2(k+1)} \right) C E = \frac{k-1}{2(k+1)} C E$.

Similar Questions

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

The electrostatic force between two charges kept in air is $F$. If $30 \%$ of the space between the charges is filled with a medium,then the electrostatic force between the charges becomes $\frac{F}{2.56}$. The dielectric constant of the medium is

$A$ 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module