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Q. A parallel plate capacitor is first connected to DC source. It is then disconnected and then immersed in a liquid dielectric. Then
(1) the capacity increases
(2) the liquid level between the plates increases
(3) the potential on the plates will decrease
(4) the liquid level will remain the same as that outside the plates

BHUBHU 2007

Solution:

As the capacitor is immersed in a liquid dielectric, its capacity increases and thus the energy decreases by $ Q^{2}/2C $ . As the energy is decreased the liquid is sucked in between the plates thus compensating the increase in potential energy in the gravitational field. In the dielectric system as there is an increase in the value of capacitance, the value of potential on the plates will decrease.