A solid sphere of radius R has a charge Q dis | vedclass

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(B) According to Gauss's law,the electric field $E$ at a distance $r$ from the center of a spherically symmetric charge distribution is given by $E(4\

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$2$

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(B) According to Gauss's law,the electric field $E$ at a distance $r$ from the center of a spherically symmetric charge distribution is given by $E(4\pi r^2) = \frac{q_{enclosed}}{\epsilon_0}$.The enclosed charge $q(r)$ within a sphere of radius $r$ is $\int_0^r \rho(r') 4\pi r'^2 dr' = \int_0^r \kappa r'^a 4\pi r'^2 dr' = \frac{4\pi \kappa r^{a+3}}{a+3}$.Thus,the electric field is $E(r) = \frac{1}{4\pi \epsilon_0 r^2} \cdot \frac{4\pi \kappa r^{a+3}}{a+3} = \frac{\kappa r^{a+1}}{\epsilon_0(a+3)}$.Given that $E(r = R/2) = \frac{1}{8} E(r = R)$,we substitute the expression for $E(r)$:$\frac{\kappa (R/2)^{a+1}}{\epsilon_0(a+3)} = \frac{1}{8} \cdot \frac{\kappa R^{a+1}}{\epsilon_0(a+3)}$.Simplifying this,we get $(1/2)^{a+1} = 1/8$.Since $1/8 = (1/2)^3$,we have $a+1 = 3$,which gives $a = 2$.

$A$ solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho = \kappa r^a$,where $\kappa$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r = \frac{R}{2}$ is $\frac{1}{8}$ times that at $r = R$,find the value of $a$.

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