Q.(a) State Gauss Law in electrostatics. Obtain an expression for Electric field due to an infinitely long charged wire. OR

(a) Gauss's law applied to a cylindrical Gaussian surface around an infinite charged wire gives $E=\lambda/(2\pi\varepsilon_0r)$; (b) balancing two cells in turn on the same potentiometer wire gives their emf ratio directly as the ratio of balancing lengths. Both alternatives answered below.

(a) Gauss's law and field due to an infinitely long charged wire

Gauss's law. 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 an infinitely long, uniformly charged straight wire (linear charge density $\lambda$).

By symmetry, $\vec E$ at any point points radially outward (for $\lambda>0$), perpendicular to the wire, and its magnitude depends only on the perpendicular distance $r$ from the wire. Choose a Gaussian surface: a coaxial cylinder of radius $r$ and length $l$, closed at both ends.

• Flux through the two flat end caps is zero, since $\vec E$ is parallel to these surfaces (perpendicular to their area vectors is not the case — rather $\vec E$ is radial while the end-cap area vector is axial, so $\vec E\cdot d\vec A=0$ there).
• Flux through the curved lateral surface: $\vec E$ is everywhere perpendicular to this surface and of constant magnitude, so the flux is simply $E\times(2\pi rl)$.

Charge enclosed: $Q_{enc}=\lambda l$.

Applying Gauss's law:

$E(2\pi rl)=\dfrac{\lambda l}{\varepsilon_0}$

$E=\dfrac{\lambda}{2\pi\varepsilon_0r}$

So the field falls off as $1/r$ (not $1/r^2$, unlike a point charge), directed radially outward from the wire.

(b) Comparing the emf of two cells using a potentiometer

Setup. A potentiometer wire is connected in series with a driver battery, a rheostat, and a key, so a steady current flows through the wire, giving a uniform potential drop per unit length $k$ along it. The cell whose emf is to be measured is connected (positive terminal to the same end as the driver battery's positive terminal) through a galvanometer and jockey to the wire, via a two-way (or single) key selecting which cell is in circuit.

…