Question

A ring of radius $R$ is having two charges $q$ and $2q$ distributed on its two half parts. The electric potential at a point on its axis at a distance of $2\surd 2R$ from its center is.$\left(k=\frac{1}{4\pi {\epsilon }_o}\right)$

Answer 1

Step 1: Given data

1. The radius of the ring is $R$.
2. Net electric charge in the ring is $3q$.
3. The distance of the point, where the electric potential has to be measured is $x=2\surd 2R$
4. Coulomb constant, $k=\frac{1}{4\pi {\epsilon }_o}$.

Step 2: Electric potential

1. The electric potential of a point is defined as the total work done to bring a positive charge from infinity to that point.
2. The electric potential at a distance r due to a charged ring of radius R is defined by the form, $P=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{q}{\sqrt{r^2+R^2}}$.

Step 3: Diagram

Step 4: Potential at point p from the center

Now, find the distance AP,

$$\begin{gathered} AP=\sqrt{R^2+{\left(2\sqrt{2}R\right)}^2}=\sqrt{\left(R^2+8R^2\right)}\left(Since,AP=2\sqrt{2}R\right) \\ orAP=\sqrt{\left(9R^2\right)}=3R \\ orAP=3R \end{gathered}$$

Now, the potential at point p is

$$\begin{gathered} P=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{3q}{3R}=K\frac{3q}{3R} \\ orP=\frac{Kq}{R} \end{gathered}$$

Therefore, the electric potential at a point on its axis at a distance $2\surd 2R$ from its center is $\frac{Kq}{R}$.