Question

Write Gauss's law. Derive the formula for electric field due to a linear charge distribution.

Hint:

The key to using Gauss's law is choosing the right Gaussian surface. The surface should be chosen such that the electric field is either parallel or perpendicular to the surface vector \(d\vec{A}\), and the magnitude of E is constant on the parts of the surface where the flux is non-zero.

Answer 1

Step 1: Statement of Gauss's Law:
Gauss's law in electrostatics states that the total electric flux (\(\Phi_E\)) through any closed hypothetical surface (called a Gaussian surface) is equal to \(\frac{1}{\epsilon_0}\) times the net electric charge (\(q_{enc}\)) enclosed within that surface.
Mathematically, it is expressed as: \[ \Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0} \] where \(\vec{E}\) is the electric field, \(d\vec{A}\) is the differential area vector on the closed surface \(S\), and \(\epsilon_0\) 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 \(\lambda\) (charge per unit length). We want to find the electric field \(E\) at a point P at a perpendicular distance \(r\) from the wire.
\begin{enumerate} \item Symmetry and Gaussian Surface: By symmetry, the electric field \(\vec{E}\) 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 \(r\) and length \(L\), 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] {$\lambda$}; % 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] {$r$}; \draw[<->] (-1.2,-1) -- (-1.2,1) node[midway, left] {$L$}; % E-field vector \draw[->, thick, magenta] (1,0) -- (1.5,0) node[right] {$\vec{E}$}; % dA vector on curved surface \draw[->, thick, green] (1,0) -- (1.5,0) node[below left, xshift=15pt, yshift=5pt] {$d\vec{A}$}; % dA vector on top surface \draw[->, thick, green] (0,1) -- (0,1.5) node[above] {$d\vec{A}$}; \node at (0.7, 1.2) [right, magenta] {$\vec{E}$}; \draw[->, thick, magenta] (0.7, 1) -- (1.2, 1); \end{tikzpicture} \end{center} \item Calculate Electric Flux (\(\Phi_E\)): 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. \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \int_{\text{top}} \vec{E} \cdot d\vec{A} + \int_{\text{bottom}} \vec{E} \cdot d\vec{A} + \int_{\text{curved}} \vec{E} \cdot d\vec{A} \] \begin{itemize} \item For the top and bottom caps, the area vector \(d\vec{A}\) is perpendicular to the electric field \(\vec{E}\) (i.e., angle is \(90^\circ\)). So, \(\vec{E} \cdot d\vec{A} = E \, dA \cos(90^\circ) = 0\). The flux through the caps is zero. \item For the curved surface, the electric field \(\vec{E}\) is parallel to the area vector \(d\vec{A}\) at every point (angle is \(0^\circ\)). So, \(\vec{E} \cdot d\vec{A} = E \, dA \cos(0^\circ) = E \, dA\). \end{itemize} The total flux is therefore only through the curved surface: \[ \Phi_E = \int_{\text{curved}} E \, dA \] Since \(E\) is constant in magnitude on this surface, we can take it out of the integral: \[ \Phi_E = E \int_{\text{curved}} dA = E \times (\text{Area of curved surface}) = E(2\pi r L) \] \item Calculate Enclosed Charge (\(q_{enc}\)): The charge enclosed by the Gaussian surface of length \(L\) is the linear charge density multiplied by the length: \[ q_{enc} = \lambda L \] \item Apply Gauss's Law: Now we equate the flux and the enclosed charge according to Gauss's law: \[ \Phi_E = \frac{q_{enc}}{\epsilon_0} \implies E(2\pi r L) = \frac{\lambda L}{\epsilon_0} \] \item Solve for E: Canceling \(L\) from both sides, we get the expression for the electric field: \[ E = \frac{\lambda}{2\pi \epsilon_0 r} \] \end{enumerate}