Answer:

The electric flux (ϕ) through curved surface =∮Edscosθ

ϕ=∮Eds [∵θ=0;cosθ=1]

=E(2rπl) [The surface area of the curved part is ] since

E

and

ds

are right angles 2πrl to each other, the electric flux through the plane caps =0.

∴ Total flux through the Gaussian surface, ϕ=E(2πrl). The net charge enclosed by Gaussian surface is, q=λl

∴ By Gauss's law,

=E(2πrl) λl/ε0

or E= λ/2πε0r

The direction of electric field E is radially outward, if line charge is positive and inward, if the line charge is negative.

Verified answer

Answer:

Gauss law :-

It states that,For any distribution of chages,The total electric flux linked with a closed surface is 1/ε0 times of the total charge within the surface.

[tex] \sf \: Mathematically \: \: \phi_E\: = \frac{Q}{ \varepsilon_o} [/tex]

the electric field of thin infinitely long straight line of Charge with a uniform charge density ?

Consider a thin infinitely long straight wire followed by speherical gussian surface of radius 'r' with the charge 'q' at the center.

Also having linear charge density λ.

By symmetry, the field E of the line charge is directed radially outward and its magnitude is same at all points equidistant from the line charge .

As , shown on the figures, it has curved surface s1 and that ends s2 and s3 respectively.

So,Total flux linked with the gussian surface =

Total flux linked with the gussian surface = sum of flux linked with all 3 surface

(Ф=Ф1 +Ф2+Ф3)

[tex]\phi_1=\oint\limits_{s_1} \vec E .d\vec s_1=\oint\limits_{s_1}E ds_1\cos0 \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: = E \int ds_1 =Es =E \times 2\pi rl \\ \phi_2=\oint\limits_{s_2} \vec E .d\vec s_2=\oint\limits_{s_2}E ds_2\cos90 =0 \\ \phi_3=\oint\limits_{s_3} \vec E .d\vec s_3=\oint\limits_{s_3}E ds_3\cos90 =0[/tex]

Ф = E × 2πrl + 0+ 0 = E × 2πrl .....(1)

But,Gauss law states that

[tex] \phi\: = \frac{Q}{ \varepsilon_o} = \frac{λl}{\varepsilon_o} ....(2)[/tex]

Equating (1) and (2)

[tex] \sf \: E × 2πrl = \frac{λl}{\varepsilon_o} \\ \\ \implies \sf E =\frac{λ}{ 2π\varepsilon_or} \: \: \: \: [/tex]

Hope it helps you

Thank you :)