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(A) To find the electric field in the annular region of a cylindrical capacitor, we apply Gauss's Law. Consider a cylindrical Gaussian surface of radi

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Varies as $\frac{1}{r}$, where $r$ is the distance from its axis

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(A) To find the electric field in the annular region of a cylindrical capacitor, we apply Gauss's Law. Consider a cylindrical Gaussian surface of radius $r$ and length $L$ coaxial with the capacitor.According to Gauss's Law, the total electric flux $\Phi_E$ through the closed surface is $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$.Due to the cylindrical symmetry, the electric field $\vec{E}$ is radial and its magnitude is constant over the curved surface of the Gaussian cylinder.The area of the curved surface is $A = 2 \pi r L$.Thus, $E 	imes (2 \pi r L) = \frac{Q_{enclosed}}{\varepsilon_0}$.If $\lambda$ is the linear charge density (charge per unit length), then $Q_{enclosed} = \lambda L$.Substituting this into the equation: $E 	imes 2 \pi r L = \frac{\lambda L}{\varepsilon_0}$.Solving for $E$, we get $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.Therefore, the magnitude of the electric field varies as $E \propto \frac{1}{r}$.

Consider a charged cylindrical capacitor. The magnitude of the electric field $\vec{E}$ in its annular region:

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