Gauss S Law

Gauss's law in electrostatics states that the total electric flux through any closed surface is equal to the net charge enclosed within the surface divided by the permittivity of free space . Mathematically, Gauss's law is expressed as: where: - is the electric field,
- is the differential area element on the closed surface,
- is the total charge enclosed within the surface,
- is the permittivity of free space, .
Now, to find the electric field due to a plane charged plate, we consider a uniformly charged infinite plane with surface charge density . We will apply Gauss's law using a Gaussian surface in the form of a cylindrical pillbox that intersects the charged plane. The electric field produced by the charged plate is symmetric and uniform, and it points perpendicular to the surface of the plate.
1. Gauss's Law Application: The Gaussian surface consists of two faces, one above and one below the charged plane. The electric field is perpendicular to the surface of the plate and has the same magnitude on both sides of the plane. The flux through the curved surface of the pillbox is zero since the electric field is parallel to this surface. Therefore, the total flux is given by the flux through the two flat faces of the pillbox.
2. Flux Calculation: The flux through one face of the pillbox is , where is the area of the face. The total flux through the two faces is: 3. Charge Enclosed: The charge enclosed by the Gaussian surface is the surface charge density times the area of the pillbox: 4. Applying Gauss's Law: Using Gauss's law, we equate the flux to the charge enclosed divided by : 5. Solving for Electric Field: Simplifying the equation, we get the electric field due to the uniformly charged plane: Thus, the electric field produced by an infinite plane of charge with surface charge density is: Key Notes: - The electric field is directed perpendicular to the surface of the charged plane.
- The electric field due to a uniformly charged plane is constant in magnitude and does not depend on the distance from the plane.

Step 1: Statement of Gauss's Law:
Gauss's law in electrostatics states that the total electric flux ( ) through any closed hypothetical surface (called a Gaussian surface) is equal to times the net electric charge ( ) enclosed within that surface.
Mathematically, it is expressed as: where is the electric field, is the differential area vector on the closed surface , and is the permittivity of free space.

Step 2: Derivation for a Linear Charge Distribution:
Consider an infinitely long, thin, straight wire with a uniform positive linear charge density (charge per unit length). We want to find the electric field at a point P at a perpendicular distance from the wire.
\begin{enumerate} \item Symmetry and Gaussian Surface: By symmetry, the electric field must be directed radially outward from the wire, and its magnitude must be the same at all points equidistant from the wire. We choose a cylindrical Gaussian surface of radius and length , coaxial with the wire, such that point P lies on its curved surface.
\begin{center} \begin{tikzpicture}[scale=1.5] % The charged wire \draw[thick, red] (0,-1.5) -- (0,1.5) node[above] {+ + +}; \node at (0.2, 0) [right] { }; % The Gaussian cylinder \draw[blue, dashed] (1,1) arc (90:270:1 and 0.3); \draw[blue, dashed] (1,-1) -- (-1,-1); \draw[blue] (-1,-1) arc (270:360:1 and 0.3) arc (0:90:1 and 0.3); \draw[blue] (-1,1) -- (1,1); \draw[blue] (-1,-1) -- (-1,1); \draw[blue] (1,-1) -- (1,1); % Labels \draw[<->] (0,0.5) -- (1,0.5) node[midway, above] { }; \draw[<->] (-1.2,-1) -- (-1.2,1) node[midway, left] { }; % E-field vector \draw[->, thick, magenta] (1,0) -- (1.5,0) node[right] { }; % dA vector on curved surface \draw[->, thick, green] (1,0) -- (1.5,0) node[below left, xshift=15pt, yshift=5pt] { }; % dA vector on top surface \draw[->, thick, green] (0,1) -- (0,1.5) node[above] { }; \node at (0.7, 1.2) [right, magenta] { }; \draw[->, thick, magenta] (0.7, 1) -- (1.2, 1); \end{tikzpicture} \end{center} \item Calculate Electric Flux ( ): The flux integral is taken over the entire closed surface, which consists of two flat circular caps (top and bottom) and the curved cylindrical surface. \begin{itemize} \item For the top and bottom caps, the area vector is perpendicular to the electric field (i.e., angle is ). So, . The flux through the caps is zero. \item For the curved surface, the electric field is parallel to the area vector at every point (angle is ). So, . \end{itemize} The total flux is therefore only through the curved surface: Since is constant in magnitude on this surface, we can take it out of the integral: \item Calculate Enclosed Charge ( ): The charge enclosed by the Gaussian surface of length is the linear charge density multiplied by the length: \item Apply Gauss's Law: Now we equate the flux and the enclosed charge according to Gauss's law: \item Solve for E: Canceling from both sides, we get the expression for the electric field: \end{enumerate}