MP Board · Class 12 · Physics · Electric Charges and FieldsState Gauss's Law in electrostatics. Using this law, derive the expression for the electric field intensity due to an infinitely long thin straight charged wire with a uniform linear charge density $\lambda$.

Statement of Gauss's Law\nGauss's Law states that the total electric flux through any closed surface is equal to $\frac{1}{\varepsilon_0}$ times the total charge enclosed by that closed surface. Mathematically, it is expressed as:

$$\oint \vec{E} \cdot d\vec{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$\nwhere $\vec{E}$ is the electric field intensity, $d\vec{S}$ is the area element vector of the closed surface, and $\varepsilon_0$ is the permittivity of free space.

Derivation for an Infinitely Long Straight Charged Wire

1. Choice of Gaussian Surface: Consider an infinitely long straight wire having a uniform linear charge density $\lambda$ (charge per unit length). Due to cylindrical symmetry, the electric field $\vec{E}$ at any point will be directed radially outward (assuming $\lambda > 0$) and its magnitude will be the same at all points equidistant from the wire. To calculate the electric field at a perpendicular distance $r$ from the wire, we choose a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the wire.

2. Electric Flux Calculation: The cylindrical Gaussian surface consists of three parts:

   • Curved cylindrical surface (Area $S_1$)
   • Two flat circular caps at the ends (Areas $S_2$ and $S_3$)

   The total electric flux $\Phi$ through the Gaussian surface is the sum of fluxes through these three surfaces: $$\Phi = \oint_{S_1} \vec{E} \cdot d\vec{S} + \int_{S_2} \vec{E} \cdot d\vec{S} + \int_{S_3} \vec{E} \cdot d\vec{S}$|

   • For the curved surface ($S_1$), the electric field $\vec{E}$ and the area vector $d\vec{S}$ are parallel everywhere (angle $\theta = 0^\circ$). Therefore, $\vec{E} \cdot d\vec{S} = E , dS \cos(0^\circ) = E , dS$.
   • For the two flat circular ends ($S_2$ and $S_3$), the electric field $\vec{E}$ is parallel to the surfaces, while the area vectors are perpendicular to them, making the angle $\theta = 90^\circ$. Therefore, $\vec{E} \cdot d\vec{S} = E , dS \cos(90^\circ) = 0$.

   Thus, the total flux simplifies to the integral over the curved surface only: $$\Phi = \int_{S_1} E , dS = E \int_{S_1} dS$$ Since the curved surface area of a cylinder of radius $r$ and length $l$ is $2\pi r l$, we get: $$\Phi = E \cdot (2\pi r l)$$

3. Applying Gauss's Law: The total charge enclosed by this Gaussian surface is: $$q_{\text{enclosed}} = \lambda \cdot l$$

   According to Gauss's Law: $$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$ $$E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon_0}$|

4. Final Expression: Cancelling $l$ from both sides and solving for $E$: $$E = \frac{\lambda}{2\pi \varepsilon_0 r}$|

   In vector form, the electric field is given by: $$\vec{E} = \frac{\lambda}{2\pi \varepsilon_0 r} \hat{n}$$ where $\hat{n}$ is the radial unit vector pointing away from the wire.