NCERT · Class 12 · Physics · Electric Charges and FieldsQuestion 1: (a) State Gauss's Law in electrostatics. (b) Using Gauss's law, derive an expression for the electric field intensity at a distance $r$ from an infinitely long straight thin wire carrying a uniform linear charge density $\lambda$. (c) Draw a neat graph showing the variation of electric field $E$ with distance $r$ from the wire.

(a) Gauss's Law in Electrostatics

Statement: Gauss's Law states that the total electric flux ($\Phi_E$) passing through any closed hypothetical surface (called a Gaussian surface) in vacuum is equal to $\frac{1}{\varepsilon_0}$ times the net electric charge ($Q_{\text{enclosed}}$) enclosed within that surface.

$$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$$ \nWhere:

• $\vec{E}$ = Electric field vector
• $d\vec{A}$ = Area element vector
• $\varepsilon_0$ = Permittivity of free space ($8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}$)

(b) Derivation of Electric Field Due to an Infinitely Long Straight Charged Wire

1. Consideration & Gaussian Surface:

   • Consider an infinitely long, straight, uniformly charged wire with linear charge density $\lambda$ (charge per unit length).
   • To calculate the electric field at a perpendicular distance $r$ from the wire, choose a co-axial cylindrical Gaussian surface of radius $r$ and length $L$ surrounding the wire.

2. Symmetry Arguments:

   • Due to cylindrical symmetry, the electric field $\vec{E}$ is directed radially outward (for positive charge) and has the same magnitude at every point on the curved surface of the cylinder.

3. Electric Flux Calculation through the Cylinder: The cylindrical surface consists of three parts:

   • Curved Surface ($S_1$): Here, electric field $\vec{E}$ and area vector $d\vec{A}_1$ are in the same direction ($\theta = 0^\circ$).
   • Top Flat Circular Cap ($S_2$): Here, $\vec{E}$ is perpendicular to area vector $d\vec{A}_2$ ($\theta = 90^\circ$).
   • Bottom Flat Circular Cap ($S_3$): Here, $\vec{E}$ is perpendicular to area vector $d\vec{A}_3$ ($\theta = 90^\circ$).

   Total flux through the closed Gaussian surface: $$\Phi_E = \iint_{S_1} E \cdot dA_1 \cos 0^\circ + \iint_{S_2} E \cdot dA_2 \cos 90^\circ + \iint_{S_3} E \cdot dA_3 \cos 90^\circ$$ $$\Phi_E = E \iint_{S_1} dA_1 + 0 + 0$$ $$\Phi_E = E \cdot (2\pi r L) \quad \text{--- (Equation 1)}$$ (Since the curved area of cylinder of radius $r$ and length $L$ is $2\pi r L$)

4. Applying Gauss's Law:

   • Total charge enclosed inside the Gaussian cylinder of length $L$ is: $$Q_{\text{enclosed}} = \lambda \cdot L$$
   • According to Gauss's Law: $$\Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda L}{\varepsilon_0} \quad \text{--- (Equation 2)}$$

5. Equating Equation 1 and Equation 2: $$E \cdot (2\pi r L) = \frac{\lambda L}{\varepsilon_0}$$ $$E = \frac{\lambda}{2\pi \varepsilon_0 r}$$

   In vector form: $$\vec{E} = \frac{\lambda}{2\pi \varepsilon_0 r} \hat{r}$$ (where $\hat{r}$ is a unit vector radial to the wire)

(c) Graphical Representation

• From the derived formula, $E \propto \frac{1}{r}$.
• As distance $r$ increases, the electric field intensity $E$ decreases hyperbolically.
• Graph Description: A rectangular hyperbola plotted with $E$ on the Y-axis and distance $r$ on the X-axis, showing asymptotic decay towards the axes.