Chapter 16, Problem 81E

(a)

To determine

To show the potential at a point on its axis at a distance $y$ from its center is $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$.

Answer to Problem 81E

The potential at a point on its axis at a distance $y$ from its center is $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$.

Explanation of Solution

The below figure (1) shows a uniformly charged disk of radius $R$.

The electric potential at a point $P$ on the axis at a distance $y$ from its center due to this ring is,

  $dV=k\frac{\sigma 2\pi rdr}{\sqrt{y^2+r^2}}$        (I)

Here, $\sigma$ is the surface charge density, $r$ is the radius of the circular disk element, $dr$ is the small thickness of the disk, $y$ is the distance from the point to the charge, and $k$ is the Coulomb’s constant.

The potential $V$ at point $P$ due to the disc is obtained by the summation over the contributions from the entire concentric ring from $r=0$ to $r=R$.

Integrate the Equation (I),

  $V=2\pi k\sigma {\displaystyle \underset{0}{\overset{R}{\int }}\frac{rdr}{\sqrt{y^2+r^2}}}$        (II)

Conclusion:

Take $x^2=r^2+y^2$ implies $xdx=rdr$.

As $r\to 0,x\to y$ and $r\to R,x\to \sqrt{y^2+R^2}$ and substitute in Equation (II) to find the potential at a point on its axis at a distance $y$ from its center.

  $\begin{array}{c}V=2\pi k\sigma {\displaystyle \underset{y}{\overset{\sqrt{y^2+r^2}}{\int }}\frac{xdx}{x}}\\ =2\pi k\sigma {\left[x\right]}_y^{\sqrt{y^2+r^2}}\\ =\frac{2\pi kQ}{\pi R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]\\ =\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]\end{array}$

Therefore, the potential at a point on its axis at a distance $y$ from its center is $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$.

(b)

To determine

To show the potential $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$ reduces to the potential for a point charge in the limit $y>>R$.

Answer to Problem 81E

The expression for the electric potential due to the point charge is $\frac{kQ}{y}$ and shown that potential $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$ reduces to the potential for a point charge in the limit $y>>R$.

Explanation of Solution

From part (a),

The potential at a point on its axis at a distance $y$ from its center is,

  $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$

Conclusion:

If $y>>R$,

Using binomial expansion,

  $\begin{array}{c}V=\frac{2kQ}{R^2}\left[y\left(1+\frac{1}{2}\frac{R^2}{y^2}\right)-y\right]\\ =\frac{2kQ}{R^2}\left[\frac{R^2}{2y}\right]\\ =\frac{kQ}{y}\end{array}$

The expression for the electric potential due to the point charge is $\frac{kQ}{y}$ and shown that the potential $V=\frac{2kQ}{R^2}\left[{\left(R^2+y^2\right)}^{1/2}-y\right]$ reduces to the potential for a point charge in the limit $y>>R$.