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Question Number 40740 by goal last updated on 27/Jul/18

Commented by goal last updated on 27/Jul/18

$$thankyousir$$

Commented by goal last updated on 27/Jul/18

$$questionno.9pleasesolvethis$$

Commented by goal last updated on 27/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$fluxforleftplaneatadistanceafromorigin$$$${\psi }_1=\alpha a^{\frac{1}{2}}{a}^{2}cos{180}^0$$$$=-\alpha a^{\frac{5}{2}}$$$${\psi }_2=\alpha {\left(2a\right)}^{\frac{1}{2}}{a}^{2}cos{0}^{o}forrightplane$$$$=\alpha .\sqrt{2}.a^{\frac{5}{2}}$$$$totalflux=\alpha a^{\frac{5}{2}}(\sqrt{2}-1)$$$$\oint E.ds=\frac{Q}{{ϵ}_0}$$$$charge=\frac{\alpha a^{\frac{5}{2}}(\sqrt{2}-1)\times {ϵ}_0}{}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$itsok...$$

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