Question

Electric field due to infinite long straight wire of linear charge density λ varies with ⊥ distance of

point from the wire as​

Answer 1

Given:

An infinite long wire of linear charge density λ

To Find:

Variation of electric field with perpendicular distance of a point from the wire

Solution:

Imagine a cylindrical gaussian surface around the wire.

Refer to the attached diagram

Curved surface area of the cylinder = 2πrl

(where 'r' is the radius and 'l' is the length of the cylinder)

Since the electric field is radial, electric flux through both the flat ends of the cylindrical gaussian surface is zero.

⇒ Flux through the total gaussian surface = Flux through the cylindrical part

⇒ $\oint E.dScos0 = \oint EdS$

⇒ $E \oint dS$

⇒ E.2πrl

The surface includes a charge (q) = λl

Hence, according to gaussian law,

E2πrl = λl/ε₀

⇒ E    = λ/2πε₀r

Thus, Electric field due to infinite long straight wire of linear charge density λ varies with ⊥ distance of  point from the wire as ​E =  λ / 2πε₀r

Answer 2

To derive:

Expression for electric field intensity due to an infinite long straight wire with linear charge density λ varies with ⊥ distance.

Solution:

Assume a cylindrical Gaussian surface of radius "r" around the infinitely long charged wire (as shown in the diagram).

Now, since the wire is infinitely long we can assume that all the field intensity vectors are directed radially outwards.

Hence, the net flux will be only through the curved surface of the cylinder (and not the flat surfaces).

Applying GAUSS' LAW:

$\therefore \displaystyle \: \oint \: \vec{E}. \vec{ds} = \dfrac{q}{ \epsilon_{0}}$

$\implies \displaystyle \: \oint \: E\times ds \times \cos( {0}^{ \circ} ) = \dfrac{q}{ \epsilon_{0}}$

$\implies \displaystyle \: E\times \oint \: ds = \dfrac{q}{ \epsilon_{0}}$

$\implies \displaystyle \: E\times 2\pi rl= \dfrac{q}{ \epsilon_{0}}$

$\implies \displaystyle \: E= \dfrac{q}{ 2\pi rl \epsilon_{0}}$

$\implies \displaystyle \: E= \dfrac{( \frac{q}{l} )}{ 2\pi r \epsilon_{0}}$

$\implies \displaystyle \: E= \dfrac{ \lambda}{ 2\pi r \epsilon_{0}}$

$\implies \displaystyle \: E \propto \: \dfrac{1}{r}$

So, the final expression is :

$\boxed{ \displaystyle \: \bf E= \dfrac{ \lambda}{ 2\pi r \epsilon_{0}} }$