Question

What is the potential (relative to infinity) at location B, a distance h from a ring of radius a with charge –Q as shown in figure 16.94?

Short answer

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

Step-by-step solution

Write the given data

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

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

The radius of ring is \(a\).

Determine the concept of electric potential

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.

Determine the electric potential at location at B

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 \( - \frac{Q}{{4\pi {\varepsilon _0}{{\left( {{a^2} + {h^2}} \right)}^{1/2}}}}\).