Find the intensity of the electric field due to the linear charge by the
law of Gauss and prove that the electric field is inversely proportional
to the distance.
Answers
Explanation:
Gauss’ Law requires integration over a surface that encloses the charge. So, our first problem is to determine a suitable surface. A cylinder of radius a that is concentric with the z axis, as shown in Figure 5.6.1 , is maximally symmetric with the charge distribution and so is likely to yield the simplest possible analysis. At first glance, it seems that we may have a problem since the charge extends to infinity in the +z and −z directions, so it’s not clear how to enclose all of the charge. Let’s suppress that concern for a moment and simply choose a cylinder of finite length l . In principle, we can solve the problem first for this cylinder of finite size, which contains only a fraction of the charge, and then later let l→∞ to capture the rest of the charge. (In fact, we’ll find when the time comes it will not be necessary to do that, but we shall prepare for it anyway.)
Here’s Gauss’ Law:
∮SD⋅ds=Qencl(5.6.1)
where D is the electric flux density ϵE , S is a closed surface with outward-facing differential surface normal ds , and Qencl is the enclosed charge.
The first order of business is to constrain the form of D using a symmetry argument, as follows. Consider the field of a point charge q at the origin (Section 5.5):
D=r^q4πr2(5.6.2)
We can “assemble” an infinite line of charge by adding particles in pairs. One pair is added at a time, with one particle on the +z axis and the other on the −z axis, with each located an equal distance from the origin. We continue to add particle pairs in this manner until the resulting charge extends continuously to infinity in both directions. The principle of superposition indicates that the resulting field will be the sum of the fields of the particles (Section 5.2). Thus, we see that D cannot have any component in the ϕ^ direction because none of the fields of the constituent particles have a component in that direction. Similarly, we see that the magnitude of D cannot depend on ϕ because none of the fields of the constituent particles depends on ϕ and because the charge distribution is identical (“invariant”) with rotation in ϕ . Also, note that for any choice of z the distribution of charge above and below that plane of constant z is identical; therefore, D cannot be a function of z and D cannot have any component in the z^ direction. Therefore, the direction of D must be radially outward; i.e., in the ρ^ direction, as follows:
D=ρ^Dρ(ρ)(5.6.3)
Next, we observe that Qencl on the right hand side of Equation 5.6.1 is equal to ρll . Thus, we obtain
∮S[ρ^Dρ(ρ)]⋅ds=ρll(5.6.4)
The cylinder S consists of a flat top, curved side, and flat bottom. Expanding the above equation to reflect this, we obtain
ρll=∫top[ρ^Dρ(ρ)]⋅(+z^ds)+∫side[ρ^Dρ(ρ)]⋅(+ρ^ds)+∫bottom[ρ^Dρ(ρ)]⋅(−z^ds)
Examination of the dot products indicates that the integrals associated with the top and bottom surfaces must be zero. In other words, the flux through the top and bottom is zero because D is perpendicular to these surfaces. We are left with
ρll=∫side[Dρ(ρ)]ds(5.6.5)
The side surface is an open cylinder of radius ρ=a , so Dρ(ρ)=Dρ(a) , a constant over this surface. Thus:
ρll=∫side[Dρ(a)]ds=[Dρ(a)]∫sideds(5.6.6)
The remaining integral is simply the area of the side surface, which is 2πa⋅l . Solving for Dρ(a) we obtain
Dρ(a)=ρll2πal=ρl2πa(5.6.7)
Remarkably, we see Dρ(a) is independent of l , So the concern raised in the beginning of this solution – that we wouldn’t be able to enclose all of the charge – doesn’t matter.
Completing the solution, we note the result must be the same for any value of ρ (not just ρ=a ), so
D=ρ^Dρ(ρ)=ρ^ρl2πρ(5.6.8)
and since D=ϵE :
E=ρ^ρl2πϵρ(5.6.9)
This completes the solution. We have found that the electric field is directed radially away from the line charge, and decreases in magnitude in inverse proportion to distance from the line charge.