Determine the capacitance with the slab in place


A parallel-plate capacitor consists of plates of area 0.19 m2 and a separation of 1.4 cm. A battery charges the plates to the potential difference of 110 V and is then detached. A dielectric slab of thickness of around 9.5 mm and dielectric constant 7.8 is then positioned symmetrically among the plates.

(i) Determine the capacitance before the slab is inserted? (ii) Determine the capacitance with the slab in place? Determine the free charge q (in C)  (iii) Before and (iv) after the slab is inserted? Determine the magnitude (in N/C) of the electric field. (v) In the space among the plates and dielectric and (vi) in the dielectric itself? (vii) By using the slab in place, determine the potential difference (in V) across the plates? (viii) Determine how much external work is comprised in inserting the slab?

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Physics: Determine the capacitance with the slab in place
Reference No:- TGS0862387

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