To calculate the electric field strength due to an infinite conducting sheet using Gauss's Law, we proceed as follows:

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Step 1: Understanding the system

The infinite conducting sheet lies on the xy-plane, with a uniform surface charge density denoted by $\sigma$. The charge is distributed uniformly over the surface, and we need to find the electric field $\vec{E}$ at any point.

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Step 2: Key assumptions

1. Due to the symmetry of the infinite sheet, the electric field is:

   • Perpendicular to the surface (along the z-direction).
   • Uniform in magnitude, regardless of the distance from the sheet.

2. The sheet is infinite, so edge effects are negligible.

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Step 3: Applying Gauss's Law

Gauss's Law is given by: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0},$ where:

• $\oint \vec{E} \cdot d\vec{A}$ is the flux of the electric field through a Gaussian surface,
• $Q_{\text{enc}}$ is the total charge enclosed by the surface,
• $\epsilon_0$ is the permittivity of free space.

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Step 3a: Choose a Gaussian surface

To exploit the symmetry of the sheet, we use a Gaussian "pillbox" surface:

• A short cylinder (pillbox) with:
  • One circular face above the sheet.
  • The other circular face below the sheet.
  • The flat faces are parallel to the sheet and of area $A$.

Step 3b: Calculate the flux

1. The electric field $\vec{E}$ points perpendicular to the sheet.
2. The flux through the sides of the pillbox is zero (as $\vec{E}$ is parallel to the sides).
3. The flux through the two flat faces is: $\Phi = E \cdot A + E \cdot A = 2EA.$

Step 3c: Total charge enclosed

The charge enclosed by the Gaussian pillbox is: $Q_{\text{enc}} = \sigma \cdot A,$ where $\sigma = \frac{Q}{A}$ is the surface charge density.

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Step 3d: Apply Gauss’s Law

Substitute the expressions into Gauss’s Law: $2EA = \frac{\sigma A}{\epsilon_0}.$

Cancel $A$ on both sides: $2E = \frac{\sigma}{\epsilon_0}.$

Solve for $E$: $E = \frac{\sigma}{2\epsilon_0}.$

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Final Result

The electric field strength due to the infinite conducting sheet is: $\boxed{E = \frac{\sigma}{2\epsilon_0}}.$

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Direction of the Electric Field

• The field points away from the sheet if $\sigma > 0$ (positive surface charge).
• The field points toward the sheet if $\sigma < 0$ (negative surface charge).

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Do you want further explanation or examples? Here are some related questions:

1. How does the result change if the sheet is non-conducting?
2. What would the field look like at a distance $z$ from the sheet?
3. How does the symmetry of the problem simplify the application of Gauss’s Law?
4. What happens if the sheet has a finite size?
5. How does this result compare to the field of a point charge?

Tip: When solving problems with symmetry, always choose a Gaussian surface that simplifies the flux calculation.