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Two concentric spherical surfaces $P_1$ and $P_2$ enclose charges $\frac{Q}{2}$ and $4 Q$ as shown in the figure. If $\phi_1$ and $\phi_2$ are the electric fluxes linked with the surfaces $P_1$ and $P_2$ respectively then
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Verified Answer
The correct answer is:
$\phi_2=9 \phi_1$
As, flux linked with a surface
$\phi=\frac{\mathrm{Q}_{\text {enclosed }}}{\varepsilon_0}$
So flux linked with surfaces $P_1$ and $P_2$ are
$\phi_1=\frac{q}{2 \varepsilon_0}$
and $\quad \phi_2=\frac{q Q}{2 \varepsilon_0}=9\left(\frac{Q}{2 \varepsilon_0}\right)$
or $\phi_2=9 \phi_1$
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