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(B) Using Gauss's Law for a cylindrical Gaussian surface of radius $x$ and length $h$ inside the volume:$\oint \vec{E} \cdot d\vec{S} = \frac{q_{enclo

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(B) Using Gauss's Law for a cylindrical Gaussian surface of radius $x$ and length $h$ inside the volume:$\oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\varepsilon_{0}}$Since the electric field is radial and uniform on the curved surface,the flux through the curved surface is $E(2\pi x h)$. The flux through the flat ends is zero.The charge enclosed is $q_{enclosed} = ho 	imes V = ho 	imes (\pi x^2 h)$.Applying Gauss's Law:$E(2\pi x h) = \frac{ho \pi x^2 h}{\varepsilon_{0}}$$E = \frac{ho x}{2\varepsilon_{0}}$Given $x = \frac{2\varepsilon_{0}}{ho}$,substituting this into the expression for $E$:$E = \frac{ho}{2\varepsilon_{0}} 	imes \left( \frac{2\varepsilon_{0}}{ho} ight) = 1 \; V m^{-1}$.

$A$ long cylindrical volume contains a uniformly distributed charge of density $\rho \; C m^{-3}$. The electric field inside the cylindrical volume at a distance $x = \frac{2 \varepsilon_{0}}{\rho} \; m$ from its axis is $....... V m^{-1}$.

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