Two concentric spherical surfaces P_1 and P_2 | vedclass

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Question

(A) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.For

Vedclass · Accepted Answer

$\phi_2 = 9\phi_1$

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(A) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.For surface $P_1$,the enclosed charge is $Q_{	ext{enclosed}, 1} = \frac{Q}{2}$.Therefore,the flux linked with $P_1$ is $\phi_1 = \frac{Q/2}{\varepsilon_0} = \frac{Q}{2\varepsilon_0}$.For surface $P_2$,the enclosed charge is $Q_{	ext{enclosed}, 2} = \frac{Q}{2} + 4Q = \frac{9Q}{2}$.Therefore,the flux linked with $P_2$ is $\phi_2 = \frac{9Q/2}{\varepsilon_0} = \frac{9Q}{2\varepsilon_0}$.Comparing the two expressions,we get $\phi_2 = 9 	imes \left(\frac{Q}{2\varepsilon_0}ight) = 9\phi_1$.

Two concentric spherical surfaces $P_1$ and $P_2$ enclose charges $\frac{Q}{2}$ and $4Q$ 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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