Respuesta :
Answer:
The magnitude of the electric field will decrease
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d
C1=K(ϵ0A/d)
C1=KC
Where C is the capacitance with no dielectric.
The new capacitance is k times the capacitance of the capacitor without dielectric slab.
This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=KC. Inserting this into equation 2
CV = KCV1
V1 = (CV)/KC
V/K
This implies the voltage decreases when a dielectric is used within the plate.
The relationship between electric field and potential voltage is a linear one
V= Ed
Therefore the electric field will decrease
