Question

$\huge{\boxed{\boxed{\red{\mathfrak{Question}}}}}$


Derive an expression for Infinite as well as finite line of charge .

✯ Don't Spam

✯ Content Quality Answers needed

✯ Don't copy

-KeepSmile ​

Answer 1

$\Huge{\underline{\underline{\mathfrak{ Question \colon}}}}$

Derive an expression for Infinite as well as finite line of charge

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

$\boxed{\setlength{\unitlength}{1 cm} \begin{picture}(7,5) \put(2,2){\oval(1, 1.08)} \qbezier(5,2.54)(5.5, 2)(5,1.46) \put(2, 2.54){\line(1,0){3}} \put(2, 1.46){\line(1,0){3}} \put(0.3,2){\line(1,0){1.7}} \put(5.24,2){\line(1,0){1.5}}\put(2,2){\circle*{0.1}} \put(1, 2.2){\vector(0,1){1}} \put(6,2.2){\vector(0,1){1}} \put(4.3,3){\vector(0,1){1}} \put(2.7,3){\vector(0,1){1}} \put(0.25,1.6){\sf Charged}\put(0.5,1.2){\sf Line}\put(2.3,1){\sf Gaussian Cylinder}\end{picture}}$

For infinitely long wire ,

Assume a cylindrical guassian surface around an uniformly charged infinite line of wire

Guass Law

The Electric Flux in a guassian surface is $\frac{1}{\epsilon}$ times the total charge enclosed

$\implies \: \sf \: \phi = \dfrac{Q}{ \epsilon_o} - - - - - - - - - (1)$

On the guassian surface, consider three small surfaces, namely $\sf ds_1 , ds_2 \ and \ ds_3$

Let $\sf ds_1 \ and \ ds_2$ be at the bases of the cylinder

• Here, the Area Vector and Electric Field Vector would be perpendicular to eachother,the angle would be 90°

$\boxed{\begin{minipage}{8 cm} \longrightarrow \displaystyle \phi \: = \sf \oint \: \vec{E} \: . \vec{ {ds}_{1} } \: \\ \\ \longrightarrow \: \displaystyle \: \sf \: \phi \: = E \: \oint \: \: {ds}_{1} \times \: cos( {90}^{ \circ} ) \\ \\ \longrightarrow \: \boxed { \boxed{\sf \: \phi \: = 0 \: Nm {}^{2} {C}^{ - 1} }} \end{minipage}}$

• Same would be expected for the second guassian surface

Let $\sf ds_3$ lie along the height of the cylinder

• Here,the Area Vector and Electric Field Vector are parallel to eachother,the angle between them would be 0°

$\boxed{\begin{minipage}{8 cm} \longrightarrow \: \displaystyle \: \sf \: \phi \: = \oint \: \vec{E} \: . \vec{ds {}_{3}} \\ \\ \longrightarrow \: \displaystyle \: \sf \: \phi \: = \oint \: E \times {ds}_{3} \times cos(0 {}^{ \circ}) = E \: \times \oint \: {ds}_{3} \\ \\ \longrightarrow \: \boxed{ \boxed{\sf \: \phi \: = E.ds}} \end{minipage}}$

• Total Surface Area of a Cylinder is 2πrl

$\longrightarrow \: \sf \: \phi \: = E(2 \pi \: rl) - - - - - - - - - (2)$

$\rule{300}{1}$

The wire is uniformly charged , there would be equal number of field lines everywhere

• Thus,the electric field would be constant throughout the guassian surface

From relation (1) and relation (2),we get :

$\sf \: E \times 2\pi \: rl = \dfrac{q}{ \epsilon_o} \\ \\ \longmapsto \: \sf \: E = \dfrac{q}{2 \pi \: rl} \times \dfrac{1}{ \epsilon_o} \:$

But $\sf \lambda = \dfrac{Q}{l}$

$\huge{ \longmapsto \boxed{ \boxed{ \sf \: E = \dfrac{ \lambda}{2\pi \: r \: \epsilon} }}} \:$

$\sf{Here}\begin{cases}\sf{E \longrightarrow Electric \ Field}\\\sf{\phi \longrightarrow Flux }\\ \sf{ \lambda \longrightarrow Linear \ Charge \ Density} \\ \sf{r,l \longrightarrow Radius, Height \ Of \ The \ Cylinder } \\ \sf{Q \longrightarrow Charge}\end{cases}$

$\rule{300}{1}$

$\rule{300}{1}$

Answer 2

Answer:

Question.....

Derive an expression for Infinite as well as finite line of charge .

Explanation:

The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward.

• KeepSmile