To determine the electric field intensity at point $P(0, 0, 20 \text{ cm})$ due to a uniformly distributed line charge on a circular ring, we will use the following approach.

Step 1: Given Data

• Charge density: $\lambda = 20 \, \text{nC/m} = 20 \times 10^{-9} \, \text{C/m}$
• Radius of the ring: $R = 15 \, \text{cm} = 0.15 \, \text{m}$
• Position of point: $P(0, 0, z_P)$, where $z_P = 20 \, \text{cm} = 0.20 \, \text{m}$
• The ring is located in the $z = 0$ plane and its center is at the origin.

Step 2: Symmetry of the Problem

Due to the symmetry of the ring, the horizontal components of the electric field will cancel out, leaving only the z-component of the electric field.

Step 3: Electric Field at a Point on the Axis of the Ring

The formula for the electric field intensity along the axis of a uniformly charged ring is:

$E_z = \frac{1}{4\pi \epsilon_0} \cdot \frac{z_P \lambda}{(R^2 + z_P^2)^{3/2}} \cdot 2\pi R$

Where:

• $\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$ (permittivity of free space)
• $z_P$ is the distance of the point from the plane of the ring along the $z$-axis
• $\lambda$ is the linear charge density
• $R$ is the radius of the ring

Step 4: Substituting the Values

Substituting the given values into the formula:

$E_z = \frac{1}{4\pi \times 8.854 \times 10^{-12}} \cdot \frac{0.20 \times 20 \times 10^{-9}}{(0.15^2 + 0.20^2)^{3/2}} \cdot 2\pi \times 0.15$

We will now calculate the result step by step.

The electric field intensity at point $P(0, 0, 20 \, \text{cm})$ due to the uniformly charged circular ring is approximately 2168.51 N/C in the positive $z$-direction.

Would you like further details or explanations?

Here are 5 related questions to explore further:

1. How does the electric field change if the charge density is doubled?
2. What is the electric field at the center of the ring?
3. How would the electric field change if the point is moved further along the $z$-axis?
4. Can we calculate the potential at the same point due to this charged ring?
5. How would the electric field change if the radius of the ring was increased?

Tip: In problems with high symmetry, such as rings or spheres, many components of the field cancel out due to symmetry, leaving only the field along certain axes.