Question

As shown in fique what will be charge on dielectric plate if dielectric constant of material is k=5 and Q= 6W​

Answer 1

Click the upper photo.

Initially when there is vacuum between the two plates, the capacitance of the capictor is

Where, A is the area of parallel plates

Suppose that the capacitor is connected to a battery, an electric field E

is produced

now if we insert the dielectric slab of thickness t=d/2 the electric field reduce to E

Eis producedow, if we insert the dielectric slab of thickness t=

t=d2the electric field reduces to ENow, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance

(d-t) the electric field is E

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