Pratham Ashish
Last Activity: 15 Years ago
Ø
electri flux trough the hemisphere will be same as through the circular surface at its mouth , coz they shere same field lines
flux through the circular surface ,
consider a thin circular strip of width dr at a distance r from the centre
dØ = E. dA
= q /4¶ε (a^2 + r ^2 ) . 2¶r dr cosθ
= -q r ( a/ √ a^2 + r^2 ) / 2 ε (a^2 + r ^2 ) { cosθ = - a/ √ a^2 + r^2}
=- (q /2ε ) ar / ( a^2 + r^2 ) ^3/2 dr
Φ = -∫ (q /2ε ) ar / ( a^2 + r^2 ) ^3/2 dr
= (q /2ε ) a / √ a^2 + r^2 ]
= - (q /2ε ) ( 1 - 1/√2 )
flux through hemisphere = - Φ
= (q /2ε ) ( 1 - 1/√2 )