Chapter 16, Problem 78E

To determine

To show that the electric field on the axis at a distance $y$ from the center of the disk has a magnitude $\underset{\_}{E=2\pi \sigma k\left[1-\left(y/{\left(y^2+R^2\right)}^{1/2}\right)\right]}$.

Answer to Problem 78E

The electric field on the axis at a distance $y$ from the center of the disk has a magnitude $\underset{\_}{E=2\pi \sigma k\left[1-\left(y/{\left(y^2+R^2\right)}^{1/2}\right)\right]}$.

Explanation of Solution

Draw a diagram of a ring with a small circular element of charge $dQ$ .

Here, $y$ is the distance from the center of the ring and the point $p$, $a$ is the radius of the ring, $O$ is the center, $\theta$ is the angle, $r$ is the distance of the point from the small segment of ring, $d{E}_{\text{x}}$ is the $x$ component of electric field, at this point, $d{E}_{\text{y}}$ is the $y$ component of electric field, at this point, and $dE$ is the net field at this point due to the ring.

From the figure:

  $\cos \theta =\frac{y}{r}$

Write the expression for the electric field due to a point charge.

  $E=\frac{kQ}{r^2}$

Here, $E$ is the electric field, $Q$ is the charge, $k$ is Coulomb’s constant, and $r$ is the distance.

Write the $y$ component of the field for the ring.

  $d{E}_{\text{y}}=\frac{kdQ}{r^2}\cos \theta$

Replace $\cos \theta$ by $y/r$ and $r$ by $\sqrt{y^2+a^2}$.

  $d{E}_{\text{y}}=\frac{kdQ}{{\left(\sqrt{y^2+a^2}\right)}^2}\left(\frac{y}{r}\right)$

  $d{E}_{\text{y}}=\frac{ykdQ}{{\left(\sqrt{y^2+a^2}\right)}^3}$        (I)

Write the expression for the surface charge density.

$\sigma =\frac{Q}{A}$

Here, $\sigma$ is the surface charge density, $Q$ is the charge, and $A$ is the area.

Write the area of the elemental ring.

$dA=2\pi ada$

Write the charge of the elemental ring.

$dQ=\sigma dA$

Replace $dA$ by $\left(2\pi ada\right)$ in the above expression.

$dQ=\sigma \left(2\pi ada\right)$

Substitute the value of $dQ$ in equation (I).

  $d{E}_{\text{y}}=\frac{yk\sigma \left(2\pi ada\right)}{{\left(\sqrt{y^2+a^2}\right)}^3}$

Conclusion:

Write the expression for the field $dE$ due to a charge element $dQ$ at the $y$ axis.

  $d{E}_{\text{y}}=\frac{yk\sigma \left(2\pi ada\right)}{{\left(\sqrt{y^2+a^2}\right)}^3}$

Integrate both sides of the above equation from limits $0$ to $a$.

  $\begin{array}{c}{\displaystyle \int d{E}_{\text{y}}}=\frac{yk\sigma \left(2\pi ada\right)}{{\left(\sqrt{y^2+a^2}\right)}^3}\\ E={\displaystyle \underset{0}{\overset{\text{R}}{\int }}\frac{2\pi y\sigma k\left(ada\right)}{{\left(y^2+a^2\right)}^{3/2}}}\\ =2\pi y\sigma k{\displaystyle \underset{0}{\overset{\text{R}}{\int }}\frac{ada}{{\left(y^2+a^2\right)}^{3/2}}}\end{array}$

Solve the above equation to get the final answer.

  $\begin{array}{c}E=2\pi y\sigma k{\left[-\frac{1}{\sqrt{y^2+a^2}}\right]}_0^R\\ =2\pi y\sigma k\left[-\frac{1}{\sqrt{y^2+R^2}}+\frac{1}{y}\right]\\ =2\pi \sigma k\left[1-\frac{y}{{\left(y^2+R^2\right)}^{\frac{1}{2}}}\right]\end{array}$

Therefore, the electric field on the axis at a distance $y$ from the center of the disk has a magnitude $E=2\pi \sigma k\left[1-\left(y/{\left(y^2+R^2\right)}^{1/2}\right)\right]$