Physics Asked by Revanth N Rajan on October 4, 2021
Assume a perfectly spherical conductor with a spherical cavity inside it. Say we place a charge $+Q$, at the center of the cavity. It induces a charge $-Q$ on the inner wall of the cavity, and a $+Q$ charge on the outer wall of the conductor. The question is regarding the flux through the "surface" of the cavity. Would it be $+Q/ε_0$? Or would the induced $-Q$ charge be involved in the equation, bringing the total enclosed charge and the flux to zero?
Please do correct me wherever wrong.
The total flux through the surface of cavity will be 0. We take the net charge enclosed in the cavity while applying gauss law.
Answered by Danny LeBeau on October 4, 2021
You seem to have misconception . When you calculate electric field with Gauss' law ,the value of electric field intensity you get is due to all charges present both inside and outside the gaussian surface.
Electric field due to a spherical charged shell inside itself is zero anyway.
Answered by Protein on October 4, 2021
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