To calculate the electric field $(E)$ at a distance $r = 0.3 \, \text{m}$ from a large square metal sheet with uniform charge distribution, we use the formula for the electric field near an infinite sheet of charge:

$E = \frac{\sigma}{2 \epsilon_0}$

Where:

• $\sigma$ is the surface charge density of the sheet.
• $\epsilon_0$ is the permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$.

Step 1: Calculate $\sigma$ (surface charge density)

The surface charge density is given by: $\sigma = \frac{Q}{A}$ Here:

• $Q = 80 \, \text{mC} = 80 \times 10^{-3} \, \text{C}$ (net charge),
• $A = \text{Area of the sheet} = \text{Length} \times \text{Width}$, where $\text{Length} = 9 \, \text{km} = 9 \times 10^3 \, \text{m}$, and $\text{Width} = 8 \, \text{km} = 8 \times 10^3 \, \text{m}$.

$A = (9 \times 10^3) \cdot (8 \times 10^3) = 72 \times 10^6 \, \text{m}^2$

Thus: $\sigma = \frac{Q}{A} = \frac{80 \times 10^{-3}}{72 \times 10^6} = 1.11 \times 10^{-12} \, \text{C/m}^2$

Step 2: Calculate $E$ (electric field)

Now, substitute $\sigma = 1.11 \times 10^{-12} \, \text{C/m}^2$ and $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$ into the formula for $E$:

$E = \frac{\sigma}{2 \epsilon_0}$

$E = \frac{1.11 \times 10^{-12}}{2 \cdot 8.85 \times 10^{-12}} = \frac{1.11}{17.7} \times 10^0$

$E \approx 0.0627 \, \text{N/C}$

Final Answer:

The electric field $0.3 \, \text{m}$ from the square metal sheet is approximately: $E \approx 0.0627 \, \text{N/C}$

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Relative Questions:

1. How does the distance from the sheet affect the electric field in this case?
2. What assumptions are made when treating the metal sheet as "infinite"?
3. How does the result change if the net charge on the sheet is doubled?
4. What is the electric field expression for a point charge instead of a sheet?
5. How is the permittivity of free space $\epsilon_0$ derived or defined?

Tip:

When dealing with large sheets or surfaces, the distance from the sheet often doesn’t significantly affect the electric field as long as $r$ is small compared to the sheet's dimensions.