Two 10-cm-diameter charged disks face each other, 20 cm apart. Both disks are charged to psint c. te axis the cle to one stiength at (a) the midpoint between the two disks and (b) a point on the axis 5 cm from one disk?

To determine the electric field strength produced by the two charged disks, we can use the formula for the electric field due to a uniformly charged disk. The electric field \(E\) at a point along the axis of a disk with surface charge density \(\sigma\) is given by:

\[
E = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)
\]

where:
- \(\sigma\) is the surface charge density,
- \(\varepsilon_0\) is the permittivity of free space (\(\approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)),
- \(z\) is the distance from the disk along its axis,
- \(R\) is the radius of the disk.

### Given:
- Diameter of disks = 10 cm → Radius \(R = 5\) cm = 0.05 m
- Distance between disks = 20 cm → \(d = 0.20\) m
- Midpoint between disks = 10 cm from either disk
- Point 5 cm from one disk = 5 cm = 0.05 m from one disk, and 15 cm = 0.15 m from the other.

### (a) Electric field at the midpoint between the disks (10 cm from either disk):
At the midpoint:
- \(z = 0.10\) m (from either disk)

For each disk, the electric field at the midpoint (due to symmetry) will add up. Assuming both disks are positively charged (the calculation is similar for negative charges):

\[
E_{\text{disk 1}} = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{0.10}{\sqrt{0.10^2 + 0.05^2}}\right)
\]

\[
E_{\text{disk 2}} = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{0.10}{\sqrt{0.10^2 + 0.05^2}}\right)
\]

Total field at midpoint:
\[
E_{\text{total}} = E_{\text{disk 1}} + E_{\text{disk 2}}
\]

### (b) Electric field at a point 5 cm from one disk:
At this point:
- Distance from Disk 1 (near) = \(z = 0.05\) m
- Distance from Disk 2 (far) = \(z = 0.15\) m

Calculating electric fields:
\[
E_{\text{disk 1}} = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{0.05}{\sqrt{0.05^2 + 0.05^2}}\right)
\]
\[
E_{\text{disk 2}} = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{0.15}{\sqrt{0.15^2 + 0.05^2}}\right)
\]

Total electric field at this point:
\[
E_{\text{total}} = E_{\text{disk 1}} - E_{\text{disk 2}} \quad \text{(fields point in opposite directions)}
\]

### Final Notes
To compute specific numerical values, the surface charge density \(\sigma\) must be known. If you have the value for \(\sigma\), you can substitute it into the formulas to find the electric fields.

Do you want detailed calculations or further assistance with any part of this? Here are some related questions you might consider:
1. How does the distance from the disks affect the electric field strength?
2. What happens to the electric field if the disks are negatively charged?
3. How can you determine the surface charge density from electric field measurements?
4. What are the implications of electric fields in practical applications?
5. How would the electric field change if the disks were larger or smaller? 

**Tip:** Always check the signs and directions of electric fields when dealing with multiple charges!

Two 10-cm-diameter charged disks face each other, 20 cm apart. Both disks are charged. Calculate the electric field strength at (a) the midpoint between the two disks and (b) a point on the axis 5 cm from one disk.

Explore the calculation of electric field strength produced by two charged disks, with a diameter of 10 cm and spaced 20 cm apart. This solution details the electric field at the midpoint and at a point 5 cm from one disk. Ideal for high school physics students.

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