Question

A disk of radius$\mathbf{2}\mathbf{.5}\mathbf{}\mathbf{\text{cm}}$has a surface charge density of$\mathbf{5}\mathbf{.3}\mathbf{}{\mathbf{\text{μC/m}}}^{\mathbf{\text{2}}}$on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance$\mathit{z}\mathbf{}\mathbf{=}\mathbf{12}\mathbf{}\mathbf{\text{cm}}$from the disk?

Short answer

The magnitude of the electric field produced by the disk at a point on its central axis is.$6.3\times {10}^{\text{3}}\text{\hspace{0.17em}N/C}$

Step-by-step solution

The given data

• Radius of the disk,$R=2.5\text{\hspace{0.17em}cm}$
• Surface density of the upper face of the disk,$\sigma =5.3{\text{\hspace{0.17em}μC/m}}^{\text{2}}$
• Distance of the point from the disk,$z=12\text{\hspace{0.17em}cm}$

Understanding the concept of electric field 

Using the given formula of the electric field of a point due to a disk at a distance from it, we can get the required magnitude value of the field.

Formula:

The magnitude of the electric field produced by the disk at a point on its central axis, $E=\frac{\sigma }{2{\epsilon }_o}(1-\frac{z}{\sqrt{z^2+R^2}})$ (i)

where,$\sigma$= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

Calculation of the magnitude of the electric field

The magnitude of the electric field produced by the disk at a point on its central axis is given using the equation (i) and the given data as follows:

$\begin{array}{c}E=\frac{(5.3\times {10}^{-6}{\text{\hspace{0.17em}C/m}}^{\text{2}})}{2(8.99\times {10}^9\text{\hspace{0.17em}N}\cdot \frac{{\text{m}}^2}{{\text{C}}^{\text{2}}})}[1-\frac{12\text{\hspace{0.17em}cm}}{\sqrt{{\left(12\text{cm}\right)}^2+{\left(2.5\text{\hspace{0.17em}cm}\right)}^2}}]\\ =6.3\times {10}^3\text{\hspace{0.17em}N/C}\end{array}$

Hence, the value of the electric field is $6.3\times {10}^3\text{\hspace{0.17em}N/C}$.