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CBSE Class 12-science Answered

Please solve this question 
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Asked by khandanya2 | 23 Jan, 2020, 05:06: PM
answered-by-expert 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 / ε= 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, 09:39: PM
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