a parallel plate capacitor with air between the plates is charged and then disconnected from the source of emf. if the space between the plates is now filled with a dielectric, what happens to the capacitance of the capacitor and the magnitude of the electric field between the plates? capacitance (a) increases (b) increases (c) increases (d) decreases (e) decreases magnitude of the electric field increases remains constant decreases increases decreases



Answer :

As the dielectric constant K exceeds 1, the capacitance rises. Possible variation V = CQ. The p.d. between the plates lowers as C rises while Q stays the same when the battery is disconnected.

Electric field E=dV, where d is the distance between the plates and V is the p.d. The electric field also diminishes when V lowers while d stays constant. U=21CQ2 is the energy stored in a capacitor. U declines while Q stays constant and C rises.

If the plates of a parallel plate capacitor are pulled apart while it is charging, what happens?

As the plates are separated, the distance d between them increases, which causes the capacitance to decrease. As a result, the capacitor's voltage rises as Q=CV in order to keep the charge constant.

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