91影视

Assume the data listed as below-

• Let 鈥楻鈥� be the distance from a fixed point charge Q.
• Let 鈥楨鈥� be the electric potential at a point.

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.

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

Electric field produced at a distance R is given by $E=\frac{Q}{4鈭�{蔚}_0{R}^2}$鈥︹赌�..(1)

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

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

Further, put the value of E from equation (1) and solve equation of 鈥榁鈥� as below:

$V=-{鈭�}_{鈭�}^{R}E.d\mathrm{R}$

$$\begin{gathered} V=-{鈭�}_{鈭�}^R\frac{Q}{4蟺{蔚}_0{R}^2}dR \\ =-\frac{Q}{4蟺{蔚}_0}脳{\left[\frac{-1}{R}\right]}_{鈭�}^R \\ =-\frac{Q}{4蟺{蔚}_0}脳\frac{-1}{R} \end{gathered}$$

$V=\frac{Q}{4蟺{蔚}_{0}R}$鈥︹赌︹赌︹赌�(2)

Now, Potential difference of a point charge is given by

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

$V=\frac{1}{4蟺{蔚}_0}\frac{Q}{R}$鈥︹赌︹赌︹赌︹赌�.(3)

Here, $K=\frac{1}{4蟺{蔚}_0}$, 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).