Question

Using Gauss's law, deduce an expression for electric field at a point due to a uniformly charged infinite plane thin sheet.

Hint:

For an infinite plane sheet of charge: \[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \] Remember:
• Electric field is uniform.
• Electric field is independent of distance from the sheet.
• The field is always perpendicular to the plane of the sheet. A common mistake is to include the distance from the sheet in the formula. For an infinite sheet, the electric field remains constant everywhere.

Answer 1

Concept: Gauss's law is one of the fundamental laws of electrostatics. It relates the total electric flux passing through a closed surface to the total charge enclosed by that surface. The mathematical statement of Gauss's law is \[ \oint \vec{E}\cdot d\vec{S} = \frac{q_{\text{enc}}}{\varepsilon_0}, \] where
• \(\vec{E}\) is the electric field,
• \(d\vec{S}\) is the outward area vector,
• \(q_{\text{enc}}\) is the total charge enclosed by the Gaussian surface,
• \(\varepsilon_0\) is the permittivity of free space. For a uniformly charged infinite plane sheet, symmetry plays a very important role. Because the sheet is infinitely large and uniformly charged, every point on the sheet is equivalent and there is no preferred direction along the plane of the sheet. Hence, the electric field:
• is perpendicular to the plane of the sheet,
• has the same magnitude at all points equidistant from the sheet,
• has equal magnitudes on both sides of the sheet.

Step 1: Consider an infinite plane sheet having uniform surface charge density \(\sigma\).
Let \[ \sigma=\frac{\text{charge}}{\text{area}} \] be the surface charge density of the sheet. To apply Gauss's law conveniently, we choose a cylindrical Gaussian surface (often called a pill-box) of cross-sectional area \(A\), such that:
• one flat face lies above the sheet,
• the other flat face lies below the sheet,
• the curved surface is perpendicular to the electric field lines. \[ \text{Area of each flat face}=A. \]

Step 2: Calculate the electric flux through the Gaussian surface.
The electric field is perpendicular to the plane sheet and hence also perpendicular to the two flat faces of the cylinder. Therefore, the electric flux through the upper face is \[ \phi_1=EA. \] Similarly, the electric flux through the lower face is \[ \phi_2=EA. \] On the curved surface, the electric field is parallel to the surface. Hence, \[ \phi_3=0. \] Therefore, the total electric flux through the Gaussian surface is \[ \phi = EA+EA+0 = 2EA. \] Thus, \[ \oint \vec{E}\cdot d\vec{S} = 2EA. \]

Step 3: Determine the charge enclosed by the Gaussian surface.
The charge enclosed by the pill-box is equal to the charge present on the portion of the sheet enclosed by its cross-sectional area \(A\). Hence, \[ q_{\text{enc}} = \sigma A. \]

Step 4: Apply Gauss's law.
According to Gauss's law, \[ \oint \vec{E}\cdot d\vec{S} = \frac{q_{\text{enc}}}{\varepsilon_0}. \] Substituting the values, \[ 2EA = \frac{\sigma A}{\varepsilon_0}. \] Cancelling \(A\) from both sides, \[ 2E = \frac{\sigma}{\varepsilon_0}. \] Therefore, \[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \] This is the magnitude of the electric field produced by a uniformly charged infinite plane sheet. Direction of the Electric Field:
• For a positively charged sheet, the electric field is directed away from the sheet on both sides.
• For a negatively charged sheet, the electric field is directed towards the sheet on both sides. Important Observation: The expression \[ E=\frac{\sigma}{2\varepsilon_0} \] does not contain the distance from the sheet. Hence, the electric field due to an infinite plane sheet is independent of the distance from the sheet and remains constant everywhere. Therefore, \[ \boxed{ E=\frac{\sigma}{2\varepsilon_0} } \] is the required expression.