Question
Electric charge is distributed in a small volume. The electric flux passing through a total charge enclosed by a spherical surface of radius 10 cm is 10 Vm . Then what will be the flux associated with a concentric sphere of radius 20 cm ?
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Answer
According to Gauss's law, the value of total flux coming out of the surface due to charge $q$ surrounded by a closed surface is
$ \phi=\oint \overrightarrow{E} \cdot d \overrightarrow{S}=\frac{q}{\epsilon_0} $
It is clear from this that the total flux associated with a closed surface depends on the charge located inside the surface and does not depend on the size and shape of the surface
$ \phi=\oint \overrightarrow{E} \cdot d \overrightarrow{S}=\frac{q}{\epsilon_0}=20 Vm $
the charge enclosed by the surface is constant
$\frac{q}{(\epsilon_0)}=constant$
$ \phi=\oint \overrightarrow{E} \cdot d \overrightarrow{S}=\frac{q}{\epsilon_0} $
It is clear from this that the total flux associated with a closed surface depends on the charge located inside the surface and does not depend on the size and shape of the surface
$ \phi=\oint \overrightarrow{E} \cdot d \overrightarrow{S}=\frac{q}{\epsilon_0}=20 Vm $
the charge enclosed by the surface is constant
$\frac{q}{(\epsilon_0)}=constant$
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