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}