Increase.

The reactance of a capacitor is given by … (i)

where ω = angular frequency of oscillation of current, C = capacitance.

Now, the capacitance of a parallel plate capacitor is given by

(ii), where k = dielectric constant, A = area of plates,

= electric permittivity of vacuum, d = distance between plates.

For vacuum, the dielectric constant k = 1. Let the capacitance in vacuum be … (iii) (from (ii))

For any other medium, k>1. Hence, capacitance of this slab is given by … (iv), where k = dielectric constant, A = area of plates, = electric permittivity of vacuum, d = distance between plates, = capacitance in vacuum

Hence, the reactance of the capacitor in vacuum will be

… (v), where ω = angular frequency, C₀ = capacitance in vacuum

And, the reactance of the capacitor in the dielectric slab will be …(vi) (from (iv)), where k = dielectric constant of slab

Since the dielectric constant k is greater than 1, it becomes clear that X_C1 > X_C2, where X_C1 = reactance of capacitor in vacuum, X_C2 = reactance of capacitor in dielectric slab

Now, the rms value of current is given by

… (vii), where i₀ = peak value of current, V₀ = peak value of voltage, X_C = capacitive reactance

Let the rms value of current initially be i₁ and then be _i2 after insertion of dielectric slab.

Therefore, , where i₀₁ = peak value of current initially, V₀ = peak value of voltage, X_C1 = reactance of capacitor in vacuum

and , where i₀₂ = peak value of current after insertion of slab, V₀ = peak value of voltage, X_C1 = reactance of capacitor after insertion of slab

Since we found out that X_C1> X_C2, it is obvious that

⇒ i₂ > i₁.

Therefore, the rms current increases. (Ans)