Let, A be the small area of the sheet containing charge q.
using Gauss theorem, net electric flux through the cylinder enclosing the charge 'q' is -
Φ = q/εο ......(1)
But, Φ = Φ1 + Φ2 + Φ3
= EACos0° + EACos0° + 0
Φ = 2EA .......(2)
From equation (1) & (2)
2EA = q/εο
E = q/2Aεο
Also, σ = q/A
Therefore,
[ E = σ/2εο ]
Still not clear
by www.advguruji.blogspot.com
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