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This is a physics problem involving an infinite charged plane with a circular hole, as shown in the diagram. The goal is to find the electric field at a point \( P \), located at a height \( z \) above the center of the hole. Here's how to approach the solution:

### Step-by-Step Solution

1. **Electric Field of an Infinite Charged Plane:**
   - For an infinite plane with uniform surface charge density \( \sigma \), the electric field at any point above or below the plane is given by:
     \[
     E_{\text{plane}} = \frac{\sigma}{2\epsilon_0}
     \]
     where \( \epsilon_0 \) is the permittivity of free space. The direction of the electric field is perpendicular to the surface.

2. **Effect of the Hole:**
   - The hole in the plane can be treated as if there is a superposition of two fields: 
     - The field due to the complete plane (without the hole).
     - The field due to a disk of radius \( R \) with the same charge density \( -\sigma \) (opposite to that of the plane) to cancel out the field in that region.

3. **Electric Field Due to the Disk:**
   - The electric field at a height \( z \) above the center of a disk with radius \( R \) and surface charge density \( -\sigma \) is:
     \[
     E_{\text{disk}} = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)
     \]
     The negative sign indicates that the field is opposite to that of the infinite plane.

4. **Net Electric Field at Point \( P \):**
   - The net electric field \( E_{\text{net}} \) at point \( P \) (above the hole) is the sum of the field due to the infinite plane and the field due to the disk:
     \[
     E_{\text{net}} = E_{\text{plane}} + E_{\text{disk}}
     \]
     Substituting the expressions:
     \[
     E_{\text{net}} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \left(-1 + \frac{z}{\sqrt{z^2 + R^2}}\right)
     \]
     Simplify:
     \[
     E_{\text{net}} = \frac{\sigma}{2\epsilon_0} \cdot \frac{z}{\sqrt{z^2 + R^2}}
     \]

Therefore, the electric field at point \( P \) is:
\[
E_{\text{net}} = \frac{\sigma z}{2\epsilon_0 \sqrt{z^2 + R^2}}
\]

### Summary
The electric field at height \( z \) above the center of the hole in the infinite plane is:
\[
E = \frac{\sigma z}{2\epsilon_0 \sqrt{z^2 + R^2}}
\]

This expression shows how the electric field varies with height \( z \) and the radius of the hole \( R \).

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Would you like more details or have any questions?

### Related Questions:
1. What happens to the electric field when \( z \) is much larger than \( R \)?
2. How does the electric field behave as \( z \) approaches zero?
3. If the surface charge density \( \sigma \) were doubled, how would the electric field change?
4. How can we use Gauss's Law to derive the electric field of an infinite plane without a hole?
5. What is the potential difference between point \( P \) and the surface of the plane?

### Tip:
When dealing with superposition problems, breaking down the scenario into simpler, additive fields (like plane and disk) can simplify the computation significantly.

In the given figure, a circular hole with a small radius R is cut out from an infinite plane with a uniform surface charge density σ. The z-axis is perpendicular to the plane and passes through the center of the hole. Determine the electric field at point P, located at a height z above the surface of the plane.

Learn how to find the electric field at a point above an infinite plane with a circular hole using principles of electromagnetism and the superposition method. This problem involves understanding surface charge density and applying Gauss's Law, making it ideal for university-level physics students.

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