NEET Class neet Answered
The plates of a parallel plate capacitor are charged upto 200volts.a dielectric slab of thickness 4mm is inserted between it's plates .then to maintain the same potential difference between the plates of capacitor,the distance between the plates is increased by 3.2mm.the dielectric constant of dielectric slab is
Asked by qudsiasultana79 | 21 Mar, 2019, 04:25: PM
Expert Answer
if the parallel plate capacitor without any dielectric is charged , its potential V is given by , V = Q / C = Q d / (A εo ) ................(1)
where A is area of plates, d is the distance between plates, Q is charge on plates and εo is permitivity of free space.
when dielectric of 4 mm thickness increased and plates are separated by another 3.2 mm to maintain the same potential difference,
we have electric field in free space E1 = σ / εo and electric field in dielectric E2 = σ / ( εo εr ) , where σ is the charge density and
εr is relative permitivity of dielectric.
hence potential difference V = V1 + V2 = E1 ( d - 0.8) + E2 4 = ( σ / εo ) [ (d - 0.8) + ( 4/εr ) ] ......................(2)
let us substitute σ = Q/A in (2), then we have V = ( Q / (A εo ) ) [ (d - 0.8) + ( 4/εr ) ] ............................ ( 3 )
since potential differenceis maintained as 200 V with and without dieclectric, we can equate eqn.(1) and eqn.(3)
Qd / (A εo ) = ( Q / (A εo ) ) [ (d - 0.8) + ( 4/εr ) ] or d = (d - 0.8) + ( 4/εr ) or 4/εr = 0.8 or εr = 5
Answered by Thiyagarajan K | 21 Mar, 2019, 10:00: PM
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