CBSE · Class 12 · Physics · Electric Charges and FieldsState Gauss's Theorem in electrostatics. Using Gauss's theorem, derive an expression for the electric field intensity at a point due to an infinitely long straight line charge having uniform linear charge density $\lambda$.

Gauss's Theorem

Statement: Gauss's Theorem states that the total electric flux ($\Phi_E$) passing through any closed surface in vacuum or air is equal to $\frac{1}{\varepsilon_0}$ times the net total charge ($q$) enclosed within that surface. $$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0}$$\nWhere $\varepsilon_0$ is the permittivity of free space.

Derivation of Electric Field due to an Infinitely Long Straight Uniformly Charged Wire

1. Consideration and Setup

• Consider an infinitely long, thin, straight wire having a uniform linear charge density $\lambda$ (charge per unit length).
• Let $P$ be a point located at a perpendicular distance $r$ from the wire, where the electric field intensity $\vec{E}$ needs to be determined.

2. Choice of Gaussian Surface

• Due to cylindrical symmetry, we choose a coaxial right circular cylinder of radius $r$ and length $l$ as the Gaussian surface, with the charged wire acting as its axis.
• The surface consists of three parts:
  1. Top circular face ($S_1$)
  2. Bottom circular face ($S_2$)
  3. Curved cylindrical surface ($S_3$)

3. Calculation of Electric Flux

• Flux through top and bottom faces ($S_1$ and $S_2$): The electric field $\vec{E}$ is directed radially outward, perpendicular to the wire, while the area vectors $d\vec{A}$ for $S_1$ and $S_2$ are parallel to the axis of the cylinder. Thus, $\theta = 90^\circ$. $$\Phi_1 = \int_{S_1} E \cdot dA \cdot \cos 90^\circ = 0$$ $$\Phi_2 = \int_{S_2} E \cdot dA \cdot \cos 90^\circ = 0$$

• Flux through curved surface ($S_3$): The electric field $\vec{E}$ and area vector $d\vec{A}$ at every point on the curved surface are in the same direction, so $\theta = 0^\circ$. $$\Phi_3 = \int_{S_3} E \cdot dA \cdot \cos 0^\circ = E \int_{S_3} dA$$ Since the curved surface area of a cylinder of radius $r$ and length $l$ is $2\pi r l$: $$\Phi_3 = E \cdot (2\pi r l)$$

• Total Electric Flux ($\Phi_E$): $$\Phi_E = \Phi_1 + \Phi_2 + \Phi_3 = 0 + 0 + E(2\pi r l) = E(2\pi r l)$$

4. Application of Gauss's Law

• Total charge enclosed inside the Gaussian cylinder of length $l$ is: $$q = \lambda \cdot l$$
• By Gauss's Theorem: $$\Phi_E = \frac{q}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0}$$

5. Equating Equations

$$E(2\pi r l) = \frac{\lambda l}{\varepsilon_0}$$ $$E = \frac{\lambda}{2\pi \varepsilon_0 r}$$ \nMultiplying numerator and denominator by 2: $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2\lambda}{r}$$

Conclusion & Important Key Points

• Inversely Proportional: The magnitude of the electric field is inversely proportional to the distance $r$ from the wire ($E \propto \frac{1}{r}$).
• Direction: The electric field is directed radially outwards if $\lambda > 0$, and radially inwards if $\lambda < 0$.