Following operations can be performed on a ca | vedclass

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

Question

(D) Let $C$ be the initial capacitance. Action $XW$: Connecting to battery $(V=E)$ and inserting dielectric $(K)$ results in charge $Q = KCE$ and elec

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) Let $C$ be the initial capacitance. Action $XW$: Connecting to battery $(V=E)$ and inserting dielectric $(K)$ results in charge $Q = KCE$ and electric field $E_{field} = E/d$. Action $WX$: Inserting dielectric $(K)$ and then connecting to battery $(V=E)$ results in charge $Q = KCE$ and electric field $E_{field} = E/d$. Thus,$XW$ and $WX$ result in the same electric field. Action $XWY$: Connecting to battery $(Q=CE)$,inserting dielectric $(Q=KCE)$,then disconnecting battery ($Q$ remains $KCE$). Action $XYW$: Connecting to battery $(Q=CE)$,disconnecting battery $(Q=CE)$,then inserting dielectric ($Q$ remains $CE$). Since $K > 1$,charge $KCE > CE$,so $XWY$ has more charge. Energy $U = Q^2 / (2C_{final})$. For $WXY$: $U = (KCE)^2 / (2KC) = (KCE^2)/2$. For $XYW$: $U = (CE)^2 / (2KC) = (CE^2)/(2K)$. Since $K > 1$,$(KCE^2)/2 > (CE^2)/(2K)$,so $WXY$ has more energy.

Similar Questions

In which of the following cases is the stored energy maximum?

In the given circuit, if point $C$ is connected to the earth and a potential of $+2000 \, V$ is applied to point $A$, the potential at point $B$ is.....$V$.

Charge on the capacitor in the given circuit in steady state condition is :- ............... $\mu C$

$A$ circuit is shown in the figure for which $C_1=(3 \pm 0.011) \mu F$,$C_2=(5 \pm 0.01) \mu F$,and $C_3=(1 \pm 0.01) \mu F$. If $C$ is the equivalent capacitance across $AB$,then $C$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module