91影视

For the point P, in order to evaluate the differential potential due to the small component, an infinitesimally small portion is taken, thus the relation is

$d{V}_P=\frac{1}{4蟺{系}_0}\frac{dQ}{x+r}$

Now the differential charge,

$\begin{array}{r}\frac{dQ}{dr}=\frac{Q}{a}\\ dQ=\frac{Qdr}{a}\end{array}$

Substituting the above equation,

$d{V}_P=\frac{1}{4蟺{系}_0}\frac{\frac{Qdr}{a}}{x+r}$

Integrating both sides:

$\begin{array}{r}{V}_P={鈭�}_0^a鈥�\frac{1}{4蟺{系}_0}\frac{\frac{Qdr}{a}}{x+r}\\ =\frac{Q}{4蟺{系}_0}{鈭�}_0^a鈥�\frac{1}{x+r}dr\\ =\frac{Q}{4蟺{系}_{0}a}\ln 鈦�\left(\frac{x+a}{x}\right)\end{array}$

$d{V}_R=\frac{1}{4蟺{系}_0}\frac{dQ}{\sqrt{y^2+r^2}}$

$\begin{array}{r}\frac{dQ}{dr}=\frac{Q}{a}\\ dQ=\frac{Qdr}{a}\end{array}$

$d{V}_R=\frac{1}{4蟺{系}_0}\frac{\frac{Qdr}{a}}{\sqrt{y^2+r^2}}$

Integrating both side;

$\begin{array}{r}{V}_R={鈭�}_0^a鈥�\frac{1}{4蟺{系}_0}\frac{\frac{Qdr}{a}}{\sqrt{y^2+r^2}}\\ =\frac{\frac{Q}{a}}{4蟺{系}_0}{鈭�}_0^a鈥�\frac{1}{\sqrt{y^2+r^2}}dr\\ =\frac{Q}{4蟺{系}_{0}a}\ln \left(\frac{\sqrt{y^2+r^2}+a}{y}\right)\end{array}$

$\begin{array}{r}{V}_P=\frac{Q}{4蟺{系}_{0}a}\ln 鈦�\left(\frac{x+a鈭�x}{a}\right)\\ =\frac{Q}{4蟺{系}_0}\left(\ln 鈦�\left(\frac{x+a}{a}\right)鈭�\ln 鈦�\left(\frac{x}{a}\right)\right)\\ =\frac{Q}{4蟺{系}_0}\frac{1}{x}\\ =\frac{Q}{4蟺{系}_{0}x}\end{array}$

Hence, the potential at point P when x becomes greater $\frac{Q}{4蟺{系}_{0}x}$.

Now, the potential at the point R, if y is much greater than the rod鈥檚 length;

$\begin{array}{r}{V}_P=\frac{Q}{4蟺{系}_{0}a}\ln \left(\frac{\sqrt{y^2+r^2+a}}{y}\right)\\ =\frac{Q}{4蟺{系}_{0}a}\ln \left(\frac{y+a}{y}\right)\\ =\frac{Q}{4蟺{系}_0}\left(\ln 鈦�\left(\frac{y+a}{a}\right)鈭�\ln 鈦�\left(\frac{y}{a}\right)\right)\\ =\frac{Q}{4蟺{系}_0}\frac{1}{y}\\ =\frac{Q}{4蟺{系}_{0}y}\end{array}$