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(B) According to Gauss's law, the electric field inside the conducting material of the cylinder must be zero. Since the wire has a charge per unit len

Vedclass · Accepted Answer

Surface charge density on the outer surface of the cylinder is $\frac{3\lambda}{2\pi R}$.

Vedclass · Answer

(B) According to Gauss's law, the electric field inside the conducting material of the cylinder must be zero. Since the wire has a charge per unit length of $\lambda$, an induced charge of $-\lambda$ per unit length must appear on the inner surface of the cylinder. The surface charge density on the inner surface is $\sigma_{in} = \frac{-\lambda}{2\pi R}$. Since the net charge per unit length of the cylinder is $2\lambda$, the charge on the outer surface per unit length is $2\lambda - (-\lambda) = 3\lambda$. The surface charge density on the outer surface is $\sigma_{out} = \frac{3\lambda}{2\pi R}$. For a Gaussian surface of radius $r > R$, the total enclosed charge per unit length is $\lambda + 3\lambda = 4\lambda$. Applying Gauss's law: $E(2\pi r) = \frac{4\lambda}{\epsilon_0} \implies E = \frac{2\lambda}{\pi \epsilon_0 r}$. Comparing with the options, statement $B$ is correct.

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