CBSE · Class 12 · Physics · Electric Charges and Fields(a) State Gauss's Theorem in electrostatics. Using Gauss's law, derive an expression for the electric field intensity at a distance $r$ from an infinitely long straight uniformly charged thin wire with linear charge density $\lambda$. (b) An infinitely long line charge produces an electric field of magnitude $9 \times 10^4 \text{ N/C}$ at a distance of $2 \text{ cm}$. Calculate the linear charge density.

Part (a) Gauss's Theorem and Derivation

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

$$\Phi_e = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$ \nwhere $\varepsilon_0$ is the absolute permittivity of free space.

Derivation of Electric Field Intensity Due to an Infinitely Long Uniformly Charged Wire:

1. System Description:
   Consider a thin, infinitely long straight wire carrying a uniform positive charge with linear charge density $\lambda$ (charge per unit length, i.e., $\lambda = \frac{q}{L}$).
   Due to axial symmetry, the direction of the electric field $\vec{E}$ at any point is directed radially outward and depends only on the radial distance $r$ from the wire.

2. Selection of Gaussian Surface:
   We choose a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the infinitely long line charge.

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

   • Top circular plane cap ($S_1$)

   • Bottom circular plane cap ($S_2$)

   • Curved cylindrical surface ($S_3$)

   • For Plane Surfaces ($S_1$ and $S_2$): The area vector $d\vec{A}$ is normal to the surface, making an angle $\theta = 90^\circ$ with the electric field $\vec{E}$.
     $$\Phi_1 = \int_{S_1} \vec{E} \cdot d\vec{A} = \int_{S_1} E \cdot dA \cos(90^\circ) = 0$$ $$\Phi_2 = \int_{S_2} \vec{E} \cdot d\vec{A} = \int_{S_2} E \cdot dA \cos(90^\circ) = 0$$

   • For Curved Surface ($S_3$): At every point on the curved surface, the electric field $\vec{E}$ and the area vector $d\vec{A}$ are in the same direction ($\theta = 0^\circ$), and the magnitude of $E$ is constant everywhere on this surface.
     $$\Phi_3 = \int_{S_3} \vec{E} \cdot d\vec{A} = \int_{S_3} E \cdot dA \cos(0^\circ) = E \int_{S_3} dA = E \cdot (2\pi r L)$$

   Therefore, total flux through the closed Gaussian surface is:
   $$\Phi_{\text{total}} = \Phi_1 + \Phi_2 + \Phi_3 = 0 + 0 + E(2\pi r L) = E(2\pi r L)$$

4. Application of Gauss's Law:
   The charge enclosed inside the Gaussian cylinder of length $L$ is:
   $$q_{\text{enclosed}} = \lambda L$$

   According to Gauss's theorem:
   $$\Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$ $$E (2\pi r L) = \frac{\lambda L}{\varepsilon_0}$$ $$E = \frac{\lambda}{2\pi \varepsilon_0 r}$$

   Multiplying the numerator and denominator by 2:
   $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2\lambda}{r}$$ \nThis is the required expression for the electric field intensity.

Part (b) Numerical Solution

Given Data:

• Electric field intensity, $E = 9 \times 10^4 \text{ N/C}$
• Distance from the wire, $r = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$
• Electrostatic constant, $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$

Formula: $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2\lambda}{r}$$

Step-by-Step Calculation:

1. Substitute the given values into the formula:
   $$9 \times 10^4 = (9 \times 10^9) \times \frac{2\lambda}{2 \times 10^{-2}}$$

2. Simplify the equation:
   $$9 \times 10^4 = 9 \times 10^9 \times \frac{\lambda}{10^{-2}}$$ $$9 \times 10^4 = (9 \times 10^{11}) \times \lambda$$

3. Solve for linear charge density $\lambda$:
   $$\lambda = \frac{9 \times 10^4}{9 \times 10^{11}}$$ $$\lambda = 10^{-7} \text{ C/m} = 0.1 \ \mu\text{C/m}$$

Final Answer: \nThe linear charge density of the wire is $10^{-7} \text{ C/m}$ (or $0.1 \ \mu\text{C/m}$).