A half ring of radius ‘ r ’ has a linear charge density $λ$ .The potential at the centre of the half ring is

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$\frac{λ}{4 {πε}_{0} r}$

$\frac{λ}{4 π^{2} ε_{0} r}$

$\frac{λ}{4 {πε}_{0} r^{2}}$

$\frac{λ}{4 ε_{0}}$

answer is A.

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Detailed Solution

potential due to small element ‘ p ’ at the centre

$\begin{array}{l} V = \int dv = \int k \frac{λ}{r} dl = \frac{1}{4 {πε}_{0}} \frac{λ}{r} \int dl \\ dv = K \cdot λ \frac{dl}{r}; = \frac{1}{4 {πε}_{0}} \frac{λ}{r} πr \Rightarrow \frac{λ}{4 ε_{0}} \end{array}$

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