To find the magnitude of the electric field produced by a uniformly charged nonconducting disk at a point along its axis, we can use the formula derived from Gauss's law for continuous charge distributions. The electric field at a point on the axis of a disk at a distance $z$ from the center of the disk is given by:

$E(z) = \frac{\sigma}{2 \epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right)$

Where:

• $E(z)$ is the electric field at a point distance $z$ from the center of the disk along its axis.
• $\sigma$ is the surface charge density of the disk.
• $\epsilon_0$ is the permittivity of free space ($\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N}\cdot\text{m}^2)$).
• $R$ is the radius of the disk.
• $z$ is the distance from the center of the disk along the axis.

Step 1: Calculate Surface Charge Density $\sigma$

The surface charge density $\sigma$ is the charge per unit area of the disk:

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

where:

• $Q = -8.00 \, \text{nC} = -8.00 \times 10^{-9} \, \text{C}$ is the total charge on the disk.
• $A = \pi R^2$ is the area of the disk.

Given $R = 1.25 \, \text{cm} = 0.0125 \, \text{m}$, the area $A$ is:

$A = \pi (0.0125)^2 = 4.91 \times 10^{-4} \, \text{m}^2$

Thus, the surface charge density $\sigma$ is:

$\sigma = \frac{-8.00 \times 10^{-9}}{4.91 \times 10^{-4}} = -1.63 \times 10^{-5} \, \text{C/m}^2$

Step 2: Apply the Electric Field Formula

We now use the formula for the electric field. The disk has a radius $R = 0.0125 \, \text{m}$, and the point $P$ is at a distance $z = 2.50 \, \text{cm} = 0.0250 \, \text{m}$.

Substituting into the electric field equation:

$E(z) = \frac{(-1.63 \times 10^{-5})}{2 (8.85 \times 10^{-12})} \left( 1 - \frac{0.0250}{\sqrt{(0.0250)^2 + (0.0125)^2}} \right)$

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Let me know if you have any questions or want more details! In the meantime, here are five questions you could consider related to this topic:

1. How does the distance from the disk affect the electric field's strength?
2. Why do we need to consider surface charge density rather than total charge alone?
3. How does the size (radius) of the disk influence the result?
4. What happens if the charge on the disk is positive instead of negative?
5. How would the electric field change if the disk were a conductor rather than nonconducting?

Tip: When calculating fields due to continuous charge distributions, integrating over the distribution often simplifies using symmetry and known formulas.