To find the electric field strength created by two charged disks, we can use the formula for the electric field due to a single uniformly charged disk:

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

Where:

• $E$ is the electric field.
• $\sigma = \frac{Q}{A}$ is the surface charge density.
• $Q$ is the charge on the disk.
• $A$ is the area of the disk ($A = \pi R^2$).
• $\epsilon_0$ is the permittivity of free space ($\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2$).
• $R$ is the radius of the disk.
• $z$ is the distance from the center of the disk along the axis.

Given:

• Diameter of disks = 10 cm, so radius $R = 5 \, \text{cm} = 0.05 \, \text{m}$.
• Distance between disks = 20 cm, so they are positioned at $z = 0$ and $z = 0.2 \, \text{m}$.
• Charge $Q = 50 \, \text{nC} = 50 \times 10^{-9} \, \text{C}$.

Step 1: Calculate Surface Charge Density $\sigma$

$A = \pi R^2 = \pi (0.05)^2 \approx 7.854 \times 10^{-3} \, \text{m}^2$ $\sigma = \frac{Q}{A} = \frac{50 \times 10^{-9}}{7.854 \times 10^{-3}} \approx 6.366 \times 10^{-6} \, \text{C/m}^2$

Step 2: Electric Field at the Midpoint (10 cm from either disk)

At the midpoint between the disks, $z = 0.1 \, \text{m}$:

$E_{1} = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0.1}{\sqrt{0.1^2 + 0.05^2}}\right)$ $E_{1} = \frac{6.366 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \left(1 - \frac{0.1}{\sqrt{0.01 + 0.0025}}\right)$ $E_{1} \approx \frac{6.366 \times 10^{-6}}{1.77 \times 10^{-11}} \left(1 - \frac{0.1}{0.1118}\right) \approx 359,000 \, \text{N/C} \times (1 - 0.894) \approx 39,000 \, \text{N/C}$

Since both disks are positively charged, the fields due to both disks at the midpoint will add up:

$E_{\text{total}} = E_{1} + E_{2} = 2E_{1} \approx 78,000 \, \text{N/C}$

Step 3: Electric Field 5 cm from One Disk

At a point 5 cm from one of the disks, $z = 0.05 \, \text{m}$:

$E_{1} = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0.05}{\sqrt{0.05^2 + 0.05^2}}\right)$ $E_{1} = \frac{6.366 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \left(1 - \frac{0.05}{0.0707}\right)$ $E_{1} \approx 359,000 \, \text{N/C} \times (1 - 0.707) \approx 105,000 \, \text{N/C}$

Now for the second disk, 15 cm away (the field points in the opposite direction):

$E_{2} = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0.15}{\sqrt{0.15^2 + 0.05^2}}\right)$ $E_{2} \approx \frac{6.366 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \left(1 - \frac{0.15}{0.1581}\right) \approx 359,000 \, \text{N/C} \times (1 - 0.948) \approx 18,000 \, \text{N/C}$

The total electric field at that point is:

$E_{\text{total}} = E_{1} - E_{2} \approx 105,000 \, \text{N/C} - 18,000 \, \text{N/C} \approx 87,000 \, \text{N/C}$

Summary:

(a) The electric field strength at the midpoint between the two disks is approximately 78,000 N/C.

(b) The electric field strength at a point on the axis 5.0 cm from one disk is approximately 87,000 N/C.

Do you need any more details or have any questions? Here are some related questions you might consider:

1. How does the electric field strength change with distance from the disks?
2. What would happen if one of the disks were negatively charged?
3. How would the field strength vary if the disks were larger?
4. Can you calculate the potential difference between the two disks?
5. What would the effect be if the disks were not parallel?

Tip: Always keep in mind that electric fields from like charges repel while opposite charges attract, which affects the direction of the field lines.