A parallel plate capacitor is charged to a potential difference of $50\, V$. It is discharged through a resistance. After $1$ second, the potential difference between plates becomes $40 \,V$. Then
Fraction of stored energy after $1$ second is ${16}/{25}$
Potential difference between the plates after $2$ seconds will be $32\, V$
Potential difference between the plates after $2$ seconds will be $20\, V$
Both $(a)$ and $(b)$
The energy required to charge a capacitor of $5\,\mu \,F$ by connecting a $d.c$. source of $20\, kV$ is......$kJ$
A capacitor of capacity $C$ has charge $Q$ and stored energy is $W$. If the charge is increased to $2Q$, the stored energy will be
A parallel plate capacitor of capacity ${C_0}$ is charged to a potential ${V_0}$
$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$
Then ${E_1}/{E_2}$ value is
The insulated spheres of radii ${R_1}$ and ${R_2}$ having charges ${Q_1}$and ${Q_2}$ respectively are connected to each other. There is
The mean electric energy density between the plates of a charged capacitor is (here $q$= charge on the capacitor and $A$= area of the capacitor plate)