A condenser is charged to a potential difference of `120V` ,its energy is `10^(-5)` J.If the battery is there and the space between the plates is filled up with a dielectric medium `(varepsilon_(r)=5)`,its new energy is .

A parallel plate capacitor of capacitance C is charged to a potential V and then disconnected with the battery. If separation between the plates is decreased by 50% and the space between the plates is filled with a dielectric slab of dielectric constant K=10 . then

An air capacitor is charged to a potential of 150 V and then the charging battery is disconnected. If the space between the plates is then completely filled with a material of dielectric constant 10 , then the potential between the plates will become

A parallel plate condenser is charged by connected it to a battery. Without disconnected the battery, the space between the plates is completely filled with a medium of dielectric constant k . Then

A parallel plate capacitor is charged completely and then disconnected from the battery. IF the separation between the plates is reduced by 50% and the space between the plates if filled with a dielectric slab of dielectric constant 10, then the potential difference between the plates

A parallel plate air condenser is charged and then disconnected from the charging battery. Now the space between the plates is filled with a dielectric then, it electric field strength between the plates

A parallel plate air condenser of capacity 4muF is charged to a potential of 1000 V. the energy of the condenser is