Q. 55

# A parallel-plate

Given,

Area, A = 400cm2 = 400 × 10–4m2

distance between plates d = 1cm = 1× 10–3m

voltage V = 100V

Thickness t = 0.5 = 5 × 10–4m

dielectric constant, k = 5

The capacitance of the capacitor without the dielectric slab is given by

Where,

A= Plate Area

d= separation between the plates,

0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2

When the dielectric slab is inserted, the capacitance becomes

where, t is the thickness of the slab

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