To solve this problem, we need to calculate the electric field along the axis of a uniformly charged disk. The key idea is to use the formula for the electric field on the axis of a charged disk.
Given:
- Radius of the disk, [tex]\( R = 35.0 \, \text{cm} = 0.35 \, \text{m} \)[/tex]
- Surface charge density, [tex]\( \sigma = 0.0079 \, \text{C/m}^2 \)[/tex]
- Distance from the center of the disk along the axis, [tex]\( z = 50.0 \, \text{cm} = 0.50 \, \text{m} \)[/tex]
To find the electric field [tex]\( E \)[/tex] at a distance [tex]\( z \)[/tex] from the center along the axis of the disk, we use the formula:
[tex]\[ E(z) = \frac{\sigma}{2 \epsilon_0} \left[ 1 - \frac{z}{\sqrt{z^2 + R^2}} \right] \][/tex]
Where:
- [tex]\( \epsilon_0 = 8.854187817 \times 10^{-12} \, \text{F/m} \)[/tex] (permittivity of free space)
Let’s break down the calculation step by step:
1. Calculate the term [tex]\( \sqrt{z^2 + R^2} \)[/tex]:
[tex]\[
\sqrt{z^2 + R^2} = \sqrt{(0.50)^2 + (0.35)^2} = \sqrt{0.25 + 0.1225} = \sqrt{0.3725} \approx 0.61 \, \text{m}
\][/tex]
2. Calculate the fraction [tex]\( \frac{z}{\sqrt{z^2 + R^2}} \)[/tex]:
[tex]\[
\frac{z}{\sqrt{z^2 + R^2}} = \frac{0.50}{0.61} \approx 0.82
\][/tex]
3. Calculate the term inside the brackets:
[tex]\[
1 - \frac{z}{\sqrt{z^2 + R^2}} = 1 - 0.82 = 0.18
\][/tex]
4. Calculate the electric field [tex]\( E(z) \)[/tex]:
[tex]\[
E(z) = \frac{\sigma}{2 \epsilon_0} \left[ 1 - \frac{z}{\sqrt{z^2 + R^2}} \right]
\][/tex]
Substitute the values:
[tex]\[
E(z) = \frac{0.0079}{2 \times 8.854187817 \times 10^{-12}} \times 0.18
\][/tex]
[tex]\[
E(z) = \frac{0.0079}{1.7708375634 \times 10^{-11}} \times 0.18
\][/tex]
[tex]\[
E(z) \approx 4.462 \times 10^8 \times 0.18
\][/tex]
[tex]\[
E(z) \approx 8.03 \times 10^7 \, \text{N/C}
\][/tex]
Thus, the electric field on the axis of the disk at 50.0 cm from the center of the disk is approximately [tex]\( 8.03 \times 10^7 \, \text{N/C} \)[/tex].
