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Question

(A) For a charged conducting cylinder of radius $R$ and linear charge density $\lambda$,the electric field $E$ at a distance $r > R$ is given by $E =

Vedclass · Accepted Answer

$A$ potential difference appears between the two cylinders when a charge density is given to the inner cylinder.

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(A) For a charged conducting cylinder of radius $R$ and linear charge density $\lambda$,the electric field $E$ at a distance $r > R$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.When a charge is given to the inner cylinder,an electric field exists in the region between the two cylinders,resulting in a potential difference $V = \int_{R_1}^{R_2} E \, dr$.If a charge is given to the outer cylinder,the electric field inside the cavity of the outer cylinder is zero due to the properties of a hollow conductor (Gauss's Law). Thus,no potential difference exists between the cylinders.If the same charge density $\lambda$ is given to both,the inner cylinder creates a field,but the outer cylinder does not contribute to the field in the region between them. Therefore,a potential difference will still exist due to the inner cylinder.Thus,option $A$ is correct because charging the inner cylinder creates a radial electric field between the cylinders,leading to a potential difference.

$A$ long,hollow conducting cylinder is kept coaxially inside another long,hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.

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