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

$\begin{array}{l} λ = \frac{q}{l} = \frac{q}{πr}; q = πrλ \\ V = \frac{1}{4 {πε}_{0}} \frac{q}{r} = \frac{1}{4 {πε}_{0}} \frac{πrλ}{r} = \frac{λ}{4 ε_{0}} \end{array}$

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