A parallel-plate capacitor that has been first charged and then disconnected from the voltage source is submerged in the vertical position into a liquid dielectric. How does the level of the dielectric. between the plates change in the process?

A parallel-plate capacitor with vertical plates is connected to a voltage source and then submerged into a liquid dielectric. How does the level of the dielectric between the plates change in the process? Explain the change of the energy stored by the capacitor

A parallel-plate capacitor is charged from a cell and then disconnected from the cell. The separation between the plates is now double.

The dielectric filling the space between the plates of a capacitor that has been charged and then disconnected from the voltage source is removed. How should the distance between the plates be changed so that the energy stored in the capacitor remains the same? Explain the origin of the change in energy.

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

Assertion: A parallel plate capacitor is charged to a potential difference 100 V , and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased. Reason: When we insert a dielectric the plates of a capacitor, the induced tend to draw in the dielectric into the field (just as neutral). we resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.

A parallel plate capacitor is first connected to a constant voltage source. It is then disconnected and ·then immersed in a liqnid dielectric, then:

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.

Figure shows a horizontal parallel plate capacitor is lowered on a liquid surface in such a way that its lower plate is just submerged in the liquid of dielectric constant k . Find the height to which the liquid level will be raised between the plates if the capacitor plates are given a surface charge density +sigma and -sigma on its plates.

A parallel plate capacitor is charged to a certain potential difference and the battery used to charged it is disconnected. Now a dielectric slab of dimensions equal to spacing between plates is introduced between the plates. What will be the changes, if any, in the charge, potential difference, capacitance, electric field strength and the energy stored in the capacitor?