Q34P - Fundamentals Of Physics

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?

 

Step-by-Step Solution

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 1: 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, $z=12\text{\hspace{0.17em}cm}$

Step 2: 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 = distance on the central axis of the disk  

R = Radius of the disk

Step 3: 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 $6.3 \times {10}^3\text{\hspace{0.17em}N/C}$.