Q.State Gauss' theorem. With the help of this theorem, find out the electrical intensity at any nearby point due to a uniformly charged thin and long straight wire. OR Define electrical dipole moment. An electrical dipole is placed within a uniform electric field (E) and is rotated to an angle ∠θ = 180°. Find out the work done.

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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