CBSE Class 12-science Answered
Please solve this question
Asked by khandanya2 | 23 Jan, 2020, 17:06: PM
Expert Answer
Radius of capacitor plates 12 cm, but distance between plates 5 mm.
Hence Electric field E between plates is given by , E = σ/εo ( field between plates of huge area )
where σ is charge density and εo is permitivity of free space
We have , E = σ / εo = Q / ( A εo ) .....................(1)
where Q is charge on plates and A is area of plates.
Rate of change of electric field is obtained by integrating eqn.(1)
dE/dt = [ 1 / (A εo ) ] ( dQ/dt ) = [ 1 / (Aεo ) ] I ................(2)
where I is charging current
Hence , ( dE/dt ) = [ 0.15 / ( π × 144 × 10-4 × 8.854 × 10-12 ) ] = 3.745 × 1011 V/( m s )
Answered by Thiyagarajan K | 23 Jan, 2020, 21:39: PM
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