A thin hollow cylinder (opened at the both ends)of radius and length both equal to L. This object carries a uniform surface-charge density $σ .$ The electrostatic potential at the point $Ρ$ on the axis of the cylinder(as shown in the figure)comes out to be $V_{P} = \frac{σ L}{k ε_{0}} l n (1 + \sqrt{h})$ Find the value of $(k + h) .$

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answer is 4.

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Detailed Solution

$d q = 2 π L d x σ, d v = \frac{2 π L σ d x}{4 π ε_{0} \sqrt{x^{2} + L^{2}}}$

$∴ d v = \frac{σ L}{2 ε_{0}} \int_{O}^{L} \frac{d x}{\sqrt{x^{2} + L^{2}}} \Rightarrow V = \frac{σ L}{2 ε_{0}} {| \ln (X + \sqrt{X^{2} + L^{2}} |}_{O}^{L}$

$V = \frac{σ L}{2 ε_{0}} [\ln (L + L \sqrt{2}) - \ln L], V = \frac{σ L}{2 ε_{0}} \ln (1 + \sqrt{2})$

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