A very long charged solid cylinder of radius | vedclass

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(B) To find the electric field inside a charged solid cylinder, we use Gauss's Law.Consider a Gaussian surface in the form of a cylinder of radius '$x

Vedclass · Accepted Answer

$\rho \frac{x}{2 k \varepsilon_{0}}$

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(B) To find the electric field inside a charged solid cylinder, we use Gauss's Law.Consider a Gaussian surface in the form of a cylinder of radius '$x$' and length '$\ell$' coaxial with the given cylinder.The total charge enclosed by this Gaussian surface is $q_{enc} = ho 	imes V = ho (\pi x^{2} \ell)$.According to Gauss's Law, the electric flux through the Gaussian surface is $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{k \varepsilon_{0}}$.Since the electric field is radial and uniform on the curved surface of the Gaussian cylinder, the flux is $E(2 \pi x \ell)$.Equating the two expressions: $E(2 \pi x \ell) = \frac{ho \pi x^{2} \ell}{k \varepsilon_{0}}$.Solving for $E$, we get $E = \frac{ho x}{2 k \varepsilon_{0}}$.

$A$ very long charged solid cylinder of radius 'a' contains a uniform charge density $\rho$. The dielectric constant of the material of the cylinder is $k$. What will be the magnitude of the electric field at a radial distance '$x$' $(x < a)$ from the axis of the cylinder?

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