Question

Show that if you are very far from the ring$\left(Z\gg R\right)$, the potential difference is approximately equal to that of a point charge. (This is to be expected because, if you are very far away, the ring appears nearly to be a point)

Short answer

$V=\frac{Q}{4\pi {\epsilon }_{0}R}$

Step-by-step solution

Identification of the given data

Assume the data listed as below-

• Let ‘R’ be the distance from a fixed point charge Q.
• Let ‘E’ be the electric potential at a point.

Significance of the potential difference.

If you are very far away, the ring appears to be nearly a point and potential difference is determined considering the point charge.

The concept of the point charge gives the magnitude of the potential difference.

Determination of the magnitude of the potential difference.

The equation of the magnitude of the potential difference is expressed as,

Electric field produced at a distance R is given by $E=\frac{Q}{4\prod {\epsilon }_0{R}^2}$……..(1)

Electric potential at that point$V=-{\int }_{\infty }^{R}E.d\mathrm{R}$

Here, R is the distance from a fixed point charge Q.

Further, put the value of E from equation (1) and solve equation of ‘V’ as below:

$V=-{\int }_{\infty }^{R}E.d\mathrm{R}$

$$\begin{gathered} V=-{\int }_{\infty }^R\frac{Q}{4\pi {\epsilon }_0{R}^2}dR \\ =-\frac{Q}{4\pi {\epsilon }_0}\times {\left[\frac{-1}{R}\right]}_{\infty }^R \\ =-\frac{Q}{4\pi {\epsilon }_0}\times \frac{-1}{R} \end{gathered}$$

$V=\frac{Q}{4\pi {\epsilon }_{0}R}$…………(2)

Now, Potential difference of a point charge is given by

$V=K\frac{Q}{R}$

$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$…………….(3)

Here, $K=\frac{1}{4\pi {\epsilon }_0}$, Q is the point charge.

Thus, the magnitude of the potential difference is approximately equal to that of a point charge as it can be seen from equation (2) and (3).