Q.State Gauss theorem in case of electrostatics. Using theorem, find the expression for the electric field of a thin infinitely long straight wire at a distance r from its axis. The linear charge density of the charged wire is λ Cm⁻¹. (1+2) OR Write down the relation between electric field and electric potential. An electrostatic field is expressed by →E = (6î + 4ĵ + 3k̂) unit. Find the electric flux through an area of 100 units lying in the XY plane in the field. (1+2)

Gauss's theorem relates enclosed charge to flux; applying it with a coaxial cylindrical Gaussian surface around an infinite line charge gives E = λ/(2πε₀r).

Gauss's theorem: The total electric flux through any closed surface equals $1/\varepsilon_0$ times the net charge enclosed:

$\oint \vec E\cdot d\vec A=\dfrac{Q_{enc}}{\varepsilon_0}$

Field of an infinite line charge: Consider an infinitely long straight wire with uniform linear charge density λ. By symmetry, $\vec E$ is radial and has the same magnitude at every point equidistant from the wire. Choose a cylindrical Gaussian surface of radius r and length l, coaxial with the wire. Flux through the curved surface = $E\times 2\pi r l$ (the flat end-caps contribute …

Gauss's theorem states the flux through a closed surface equals $q_{enc}/\varepsilon_0$; applying it to a cylindrical Gaussian surface around a line charge gives $E=\lambda/2\pi\varepsilon_0 r$.

Gauss's theorem: The total electric flux through any closed surface (a Gaussian surface) equals $1/\varepsilon_0$ times the total charge enclosed by that surface:

$\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$

Application — infinite line charge: Consider an infinitely long straight wire with uniform linear charge density $\lambda$. By symmetry, the field $\vec E$ at any point is radial (perpendicular to the wire) and has the same magnitude at every point at a fixed perpendicular distance $r$ from the wire.

Choose a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the wire. …

Gauss' theorem states the flux through a closed surface is $q_{enc}/\varepsilon_0$; applying it to a cylindrical Gaussian surface around a long charged wire gives $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$.

Gauss' theorem: The total electric flux through any closed surface (a Gaussian surface) equals $1/\varepsilon_0$ times the total charge enclosed by that surface:

$\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$

Field due to a long straight charged wire: Consider an infinitely long thin straight wire with uniform linear charge density $\lambda$. By symmetry, the field $\vec E$ at any point is radial and has the same magnitude at every point equidistant from the wire. Choose a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the wire.

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