A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant `K= (5)/(3)` is inserted between the plates, the magnitude of the induced charge will be :

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K =(5)/(3) is inserted between the plates the magnitude of the charge will be :

A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E . If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become.

A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E . If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become.

A parallel plate capacitor has a capacitance of "5pF" ,a plate area of 20.0mm^(2) and voltage across it is 10V.A dielectric slab with dielectric constant "5" is introduced in between the plates.Find the magnitude of the induced surface charge on the slab.

Two identical parallel plate capacitors are connected in series to a battery of 100V . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

Two identical parallel plate capacitors are connected in series to a battery of 100V . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

A parallel plate capacitor having capacitance C farad is connected with a battery of emf V volts. Keeping the capacitor cannected with the battery, a dielectric slab of dielectric constant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates. Then,

A parallel plate capacitor is connected to a battery of emf V volts. Now a slab of dielectric constant k=2 is inserted between the plates of capacitor without disconnecting the battery. The electric field between the plates of capacitor after inserting the slab is E= (PV)/(2d) . Find the value of P.

A parallel plate capacitor is connecyed to a battery of emf V volts as shown. Now a slab of dielectric constant k=2 is inserted between the plates of capacitor without disconnecting the battery. The electric field between the plates of capacitor after inserting the slab is E= (PV)/(2d) . Find the value of P.