Question

Derive an expression for electric field at a point due to a charged infinity long thin conducting w i r e using gauss law

Answer 1

Answer:

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Explanation:

Gauss's law :

The law relates the flux through any closed surface and the net charge enclosed with in the surface. The law states that the total flux of the electric field E over any closed surface is equal to

ε

0

1

times the net charge enclosed by the surface.

ϕ=

ε

0

q

This closed imaginary surface is called Gaussian surface. Gauss's law tells us that the flux of E through a closed surface S depends only on the value of net charge inside the surface and not on the location of the charges. Charges outside the surface will not contribute to flux.

Electric field due to an infinitely long straight uniformly charged wire :

Consider an uniformly charged wire of infinite length having a constant linear charge density λ (Charge per unit length). Let P be a point at a distance r from the wire and E be the electric field at the point P. A cylinder of length l, radius r, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area ds on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward.

E

and

ds

are along the same direction.

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)

ε

0

λl

or E=

2πε

0

r

λ

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

Answer 2

Answer:

Using Gauss s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. The electric field components in the following figure are `E_(x) = alphax, E_(y) = 0, E_(z) = 0, \" in which \" alpha = 400 N//C` m

Explanation:

Gauss's law :

The law relates the flux through any closed surface and the net charge enclosed with in the surface. The law states that the total flux of the electric field E over any closed surface is equal to ε0​1​ times the net charge enclosed by the surface.

ϕ=ε0​q​

This closed imaginary surface is called Gaussian surface. Gauss's law tells us that the flux of E through a closed surface S depends only on the value of net charge inside the surface and not on the location of the charges. Charges outside the surface will not contribute to flux.

Electric field due to an infinitely long straight uniformly charged wire :

Consider an uniformly charged wire of infinite length having a constant linear charge density λ (Charge per unit length). Let P be a point at a distance r from the wire and E be the electric field at the point P. A cylinder of length l, radius r, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area ds on the Gaussian surface. By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. E

and ds

are along the same direction.

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)ε0​λl​ or E=2πε0​rλ​

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