A solid metal sphere has a charge of +3Q. It | vedclass

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(B) According to Gauss's Law,the electric field at a distance $R$ from the center of a spherical charge distribution is given by $E = \frac{k q_{enclo

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$\frac{3Q}{4\pi \varepsilon_0 R^2}$

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(B) According to Gauss's Law,the electric field at a distance $R$ from the center of a spherical charge distribution is given by $E = \frac{k q_{enclosed}}{R^2}$.For the region $a The charge enclosed by this Gaussian surface is only the charge on the solid metal sphere,which is $+3Q$.The conducting shell (charge $-Q$) is outside the Gaussian surface,so it contributes zero to the electric field at distance $R$ inside the shell.Therefore,the electric field $E$ is given by:$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{3Q}{R^2} = \frac{3Q}{4\pi \varepsilon_0 R^2}$.

$A$ solid metal sphere has a charge of $+3Q$. It is concentric with a conducting spherical shell having a charge of $-Q$. The radius of the sphere is $a$ and the radius of the shell is $b$ $(b > a)$. The electric field at a distance $R$ from the center $(a < R < b)$ is .......

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