Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

State Gauss law in electrostatics. Derive an of electric field expression due an infinitely long straight uniformly charged wire. Draw necessary diagrams.

Answer
VerifiedVerified
139.8k+ views
1 likes
like imagedislike image
Hint: In electrostatics, the Gauss’s law states that the total electric flux over any closed surface is 1/1εoεo times the total charge that is enclosed by the surface. To find the electric field due to an infinitely long current carrying wire, we need to consider a cylindrical Gaussian surface around it.

Complete Step by Step Solution: The Gauss’s law in electrostatics shows us that there is an important relationship that exists between the total electric flux over any closed surface and the total charge that is enclosed by the surface. It states that the total electric flux over any closed surface is 1/1εoεo times the total charge that is enclosed by the surface where εo is the permittivity in free space.
The mathematical form of the Gauss’s law is,
SEdS=Qencεo
To derive the expression of the electric field due to an infinitely long charged wire, we consider the following diagram,

Here we have considered the Gaussian surface as a cylinder that is surrounding the wire which has a surface charge density λ. This cylinder has a radius R and a length L. Let us consider a point P lying on the curved surface of the cylinder. So we need to find the electric field due to the wire at the P.
So, from the Gauss’s law we can write,
SEdS=Qencεo
In the cylinder there are 3 surfaces present as shown in the figure. So we need to find Gauss's law for all the three surfaces S1, S2 and S3.
So we can break the L.H.S. of the integration into three parts as,
S1EdS+S2EdS+S3EdS=Qencεo
Now the direction of the electric field from the wire is radially outwards. So the electric field is perpendicular on the surface S2 and is parallel to the surfaces S1 and S3. So the angle between the electric field and the surface vectors are 90 for S1 and S3, and 0 for S2as the surface vectors are perpendicular to the surfaces. Therefore, we can write, the dot products in form of the magnitudes as,
S1EdScos90+S2EdScos0+S3EdScos90=Qencεo
The value of cos0=1 and cos90=0
So the first and the third term of the equation vanish.
Therefore, we get
S2EdS=Qencεo
Now since the electric field is constant we can take it out of the integration. So,
ES2dS=Qencεo
The surface integration is over the curved surface of the cylinder. The surface area of the curved surface S2 is S2dS=2πRL
And as the surface charge density of the wire is λ, so the total charge of the wire for L length is,
Qenc=λL
So substituting the values,
E×2πRL=λLεo
The L gets cancelled from both the sides and then taking all the terms except E from the LHS to the RHS we get
E=λ2πεoR
This is the magnitude of the electric field. The direction of the electric field is radially outwards, so we can write it as
E=λ2πεoRR^

Note: From the formula for the electric field, we can see that it is independent of the length of the wire but depends on the charge density. In case of 3D we need to replace the linear charge density of the wire with its volume charge density.