A ring of radius R has charges ‘q’ and ‘2q’ on its, two half parts respectively. The electric potential at a point on its axis at a distance $2 \sqrt{2} R$ from its centre is

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$\frac{1}{4 {πε}_{0}} \frac{q}{3 R}$

$\frac{1}{4 {πε}_{0}} \frac{q}{R}$

$\frac{1}{4 {πε}_{0}} \frac{3 q}{R}$

answer is B.

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Detailed Solution

$\begin{array}{l} V = \frac{1}{4 {πε}_{0}} (\frac{q + 2 q}{\sqrt{R^{2} + (2 \sqrt{2} R)^{2}}}) = \frac{1}{4 {πε}_{0}} \frac{3 q}{3 R} \\ = \frac{1}{4 {πε}_{0}} \frac{q}{R} \end{array}$

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