A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

A dielectric slab is inserted between the plates of an isolatted charged capacitor. Which of the following quantities will remain the same?

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q' .

A parallel - plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected . A slab of dielectric constant K is then inserted between the plate of the capacitor so as to fill the space between the plate .Find the work done on the system in the process of inserting the slab.

A parallel plate capacitor is charged by a battery after some time, the battery is disconnected and a dielectric slab with its thicnkess equal to the plate so reparation is insected between the plates. How will (i) the capacitance of the capacitor (ii) potential difference between the plates & (iii) the energy sotred in the capacitor the affected ? Q_(0) -charge V_(0) - potential difference, C_(0) - capacitance, E_(0) electric field. U_(0) - energy spred, before the dielectric slab is inserted. Q_(0)=C_(0) V_(0), (V_(0))/(d), U_(0)=(1)/(2)C_(0)V_(0)^(2)?

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates (figure 31-Q2).The capacitance now becomes.

A parallel plate condenser has a uniform electric field E (V//m) in the space between the plates. If the distance between the plates is d (m) and area of each plate is A (m^2) the energy (joules) stored in the condenser is:

Find the force of attraction between the plates of a parallel plate capacitor.

Choose the correct statements from the following for the given situation. The dielectric can be inserted into the plates of a capacitor when (i) The capacitor is connected to the battery (ii) the capacitor is disconnected from the battery (iii) the capacitor is discharging and plates are removed (iv) the capacitor is charging without the plates

A parallel plate capacitor of capacitance 100 microfarad is connected a power supply of 200 V. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. (a) Find the extra charge flown through the power supply and the work done by the supply. (b) Find the change in the electrostatic energy of the electric field in the capacitor.

A pallel- plate capacitor has plate area 100cm^2 and plate separation 1.0cm.A glass plate (dielectric constant 4.0)are inserted one over the other the fill the space between the plate of the capacitor.Find the new capacitance.