Answer any one question. a. State and explain Gauss's law in electrostatics and use it to determine the electric field intensity due to a line charge. b. What is capacitance of a capacitor? Derive a…

a. Gauss's Law in Electrostatics and Electric Field Due to a Line Charge

Statement and Explanation

Gauss's law in electrostatics states that the total electric flux through any closed surface is directly proportional to the net charge enclosed by that surface. Mathematically, it is expressed as:

$${\Phi }_E={\oint }_S\mathbf{E}\cdot d\mathbf{A}=\frac{Q_{\text{enc}}}{{ϵ}_0}$$

where:

• ${\Phi }_E$ = Electric flux (Nm²/C),
• $\mathbf{E}$ = Electric field (N/C),
• $d\mathbf{A}$ = Infinitesimal area vector (m²),
• $Q_{\text{enc}}$ = Charge enclosed (C),
• ${ϵ}_0$ = Permittivity of free space ($8.85\times 10^{-12}{\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$).

This law is derived from Coulomb’s law and is useful for calculating electric fields in symmetric charge distributions.

Electric Field Due to an Infinitely Long Line Charge

Consider an infinitely long line charge with a linear charge density $\lambda$ (C/m). To find the electric field at a distance $r$ from the line, we use a cylindrical Gaussian surface of radius $r$ and length $L$.

1. Electric Flux Calculation:

   • The electric field $\mathbf{E}$ is radial and symmetric, so its magnitude is constant on the curved surface.
   • Flux through the curved surface: $${\Phi }_E=E\cdot (2\pi rL)$$
   • Flux through the flat ends is zero because $\mathbf{E}$ is perpendicular to $d\mathbf{A}$.

2. Charge Enclosed: $$Q_{\text{enc}}=\lambda L$$

3. Applying Gauss’s Law: $$E\cdot 2\pi rL=\frac{\lambda L}{{ϵ}_0}$$

4. Solving for $E$: $$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$$

Conclusion: The electric field due to an infinitely long line charge at a distance $r$ is inversely proportional to $r$ and depends on the linear charge density $\lambda$.

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b. Capacitance and Energy Stored in a Capacitor

Capacitance of a Capacitor

The capacitance ($C$) of a capacitor is defined as the ratio of the charge stored ($Q$) on one of its plates to the potential difference ($V$) between the plates:

$$C=\frac{Q}{V}$$

Unit: Farad (F), where $1\text{F}=1\text{C/V}$.

Derivation of Energy Stored in a Capacitor

When a capacitor is charged to a potential difference $V$, work is done to separate charges, and this work is stored as electrical potential energy ($U$).

1. Work Done in Charging:

   • Suppose the capacitor is charged from $0$ to $Q$.
   • At any instant, if the charge on the capacitor is $q$, the potential difference is $v=\frac{q}{C}$.
   • The work done ($dW$) to add a small charge $dq$ is: $$dW=vdq=\frac{q}{C}dq$$
   • Total work done (energy stored) is the integral from $0$ to $Q$: $$U={\int }_0^Q\frac{q}{C}dq=\frac{1}{C}{\int }_0^{Q}qdq=\frac{Q^2}{2C}$$

2. Alternative Expression in Terms of $V$: Since $Q=CV$, substituting gives: $$U=\frac{(CV{)}^2}{2C}=\frac{1}{2}C{V}^2$$

Final Relation: The energy stored in a capacitor when charged to a potential difference $V$ is: $$U=\frac{1}{2}C{V}^2\text{or}U=\frac{Q^2}{2C}$$