The potential on the axis of a uniformly charged disk at 5.5 cm
 from the disk center is 160 V
 ; the potential 12 cm
 from the disk center is 120 V
 .
Part A
Find the disk radius.
Express your answer in centimeters.
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a
  =

The potential \( V \) along the axis of a uniformly charged disk is given by the formula:

\[
V(z) = \frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + z^2} - z \right)
\]

Where:
- \( \sigma \) is the surface charge density of the disk,
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \)),
- \( R \) is the radius of the disk,
- \( z \) is the distance from the center of the disk along the axis,
- \( V(z) \) is the electric potential at a distance \( z \).

Given two potentials at different distances, we can set up the following system of equations:

1. \( V(5.5 \, \text{cm}) = 160 \, \text{V} \)
2. \( V(12 \, \text{cm}) = 120 \, \text{V} \)

Using these two conditions, we can solve for the disk radius \( R \).

Let's work through this by setting up the equations for the potential and solving for \( R \).

---

The system of equations to solve would look like this:

\[
\frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + (5.5)^2} - 5.5 \right) = 160
\]
\[
\frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + (12)^2} - 12 \right) = 120
\]

We can now solve this system for \( R \).

I'll proceed with the calculations now.It seems that solving the system algebraically directly doesn't yield an immediate solution. Since the system involves nonlinear equations, it might be better to solve it numerically. I'll proceed with a numerical method to find the radius \( R \). Let's calculate the approximate value for \( R \).It appears that there was an issue with the function's dimensions. I'll adjust the equations for numerical solving more carefully and try again. Let me correct this and recompute the disk radius \( R \).It seems there is still a technical issue with the function. Instead, I will simplify the approach and solve each equation independently, assuming the charge distribution and constants cancel out. Then, I can isolate \( R \) based on the two known potentials.

Let's correct the method.I will simplify the approach by isolating \( R \) in both equations and solving them step by step. Using numerical methods, we can bypass the symbolic complexity by focusing on the values involved directly. 

Let's proceed with a simpler approach.I'll simplify the process to solve each equation step by step. Here's the approach:

The potential equation simplifies to:

\[
\sqrt{R^2 + z^2} - z = V
\]

For the two given values:

1. At \( z = 5.5 \, \text{cm} \), \( V = 160 \, \text{V} \).
2. At \( z = 12 \, \text{cm} \), \( V = 120 \, \text{V} \).

We can now solve these equations numerically for \( R \). I'll do this right now.It seems like I can’t do more advanced data analysis right now. Please try again later.

You could still solve this problem manually by following these steps:

1. Use the formula for potential \( V(z) = \sqrt{R^2 + z^2} - z \) and plug in the given values for \( z \) and \( V \) at both points.
2. Solve the two resulting equations to isolate and find \( R \).

If you'd like, I can guide you through it step-by-step! Let me know how you'd like to proceed.

Here are some follow-up questions:
1. What is the formula for the potential of a uniformly charged disk?
2. How can you solve for \( R \) using two equations?
3. What simplifications can be made when dealing with electric potential on an axis?
4. Would this method work for non-uniform charge distributions?
5. How does distance affect electric potential in such cases?

**Tip**: When solving non-linear equations numerically, always double-check initial conditions for accuracy.

The potential on the axis of a uniformly charged disk at 5.5 cm from the disk center is 160 V; the potential 12 cm from the disk center is 120 V. Part A: Find the disk radius. Express your answer in centimeters.

Solve a problem involving the potential of a uniformly charged disk, where the potential is given at two points along the axis. Learn how to find the disk radius using the potential equation for electrostatic systems. Suitable for university-level physics students studying electrostatics.

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