After charging a capacitor,the battery is rem | vedclass

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

Question

(A) When the battery is removed,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab o

Vedclass · Accepted Answer

The potential difference between the plates and the energy stored will decrease,but the charge on the plates will remain the same.

Vedclass · Answer

(A) When the battery is removed,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab of dielectric constant $K > 1$ is inserted between the plates,the capacitance increases according to the formula $C' = KC$.Since $Q$ is constant and $C' = KC$,the new potential difference $V' = Q/C' = Q/(KC) = V/K$. Thus,the potential difference decreases.The energy stored in the capacitor is given by $U = Q^2 / (2C)$. Since $C$ increases,the energy stored $U' = Q^2 / (2C') = U/K$ decreases.Therefore,the potential difference and the energy stored decrease,while the charge remains the same.

After charging a capacitor,the battery is removed. Now,by placing a dielectric slab between the plates :-

Similar Questions

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

$A$ parallel plate air-filled capacitor of capacitance $C_1$ has plate area $A$ and the distance between the plates $d$. When a metal sheet of thickness $\frac{d}{2}$ and of the same area $A$ is introduced between the plates,its capacitance becomes $C_2$. The ratio $C_2: C_1$ is

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module