91Ó°ÊÓ

The charge on the ring is\(q = - Q\).

The distance of location B from center of ring is\(h\).

The radius of ring is \(a\).

The electric potential is the effect of charged particle at some distance from the charged particle. The charge always moves from higher effect to lower effect.

The distance of the location B from the ring is given as:

\(d = \sqrt {{a^2} + {h^2}} \)

The electric potential at location B due to ring is given as:

\(V = \frac{q}{{4\pi {\varepsilon _0}d}}\)

Substitute all the values in the above equation.

\(\begin{array}{c}V = \frac{{\left( { - Q} \right)}}{{4\pi {\varepsilon _0}\left( {\sqrt {{a^2} + {h^2}} } \right)}}\\V = - \frac{Q}{{4\pi {\varepsilon _0}{{\left( {{a^2} + {h^2}} \right)}^{1/2}}}}\end{array}\)

Therefore, the electric potential at location B due to ring is \( - \frac{Q}{{4\pi {\varepsilon _0}{{\left( {{a^2} + {h^2}} \right)}^{1/2}}}}\).