For a uniform line charge of 8 nc/m lying along the z axis, the electric field at a radius of 3 m from the uniform line is given by :
47.9 aρ V/m
The field of an infinite line charge falls as 1/ρ and points radially outward :
\(\vec{E}=\dfrac{\rho_{L}}{2\pi\varepsilon_{0}\rho}\,\vec{a}_{\rho}\)
Substituting the given values, with \(\rho_{L}=8\times10^{-9}\) C/m and \(\rho=3\) m:
\(2\pi\varepsilon_{0}\rho=2\pi\times8.854\times10^{-12}\times3=1.669\times10^{-10}\)
\(E=\dfrac{8\times10^{-9}}{1.669\times10^{-10}}=47.9\ \text{V/m}\)
so \(\vec{E}=47.9\,\vec{a}_{\rho}\) V/m — option 3.
A faster route using the standard constant. Since \(1/2\pi\varepsilon_{0}=17.98\times10^{9}\), the formula becomes
\(E=\dfrac{17.98\times10^{9}\times8\times10^{-9}}{3}=\dfrac{143.8}{3}=47.9\ \text{V/m}\)
Where the formula comes from. Apply Gauss's law to a cylinder of radius \(\rho\) and length L coaxial with the line. By symmetry E is radial and constant over the curved surface, and the flat ends contribute nothing since E is parallel to them:
\(E\left(2\pi\rho L\right)=\dfrac{\rho_{L}L}{\varepsilon_{0}}\quad\Rightarrow\quad E=\dfrac{\rho_{L}}{2\pi\varepsilon_{0}\rho}\)
The L cancels, which is what makes the result independent of how much of the line is enclosed.
Why the exponent is 1 and not 2. A point charge spreads its flux over a sphere of area \(4\pi r^{2}\), giving \(1/r^{2}\). An infinite line spreads it over a cylinder, whose area grows only as \(2\pi\rho L\) — one power of the distance — so the field falls as \(1/\rho\). By the same argument an infinite sheet gives a field that does not fall off at all.
Options 1 and 3 differ only by transposed digits, 49.7 against 47.9, so the arithmetic has to be carried through rather than estimated. Option 2 is roughly double the correct value, which is what results from omitting the factor of 2 in the denominator — that is, from using the point-charge constant \(1/4\pi\varepsilon_{0}\) in the wrong place.
Hence, E = 47.9 aρ V/m.
Similar Questions
Match the following lists in terms of relative permittivity :
List – I List – II a. Air i. 25 b. Alcohol (Ethyl) ii. 4 to 7 c. Glass iii. 80 d. Fresh water iv. 1 Given below are two statements :
Statement I : When an electron is brought near the surface of a metal in any dielectric medium, an image charge of opposite polarity is included inside the metal at the same distance from the surface as that of the electron in the dielectric medium.
Statement II : In the case of metal-semiconductor junction, an electron in the semiconductor also creates an image charge in the metal which decreases the energy barrier height.
In the light of the above statements, choose the correct answer from the options given below :
Important Questions from Electrostatics
A uniform ring charge of radius $R$ carries a total charge $Q$. Which one of the following options correctly quantifies the magnitude of the force on a point charge of strength $q$ kept at the center of the ring?
($\epsilon$ is the permittivity of the medium)
The current in a coil of inductance 5 H changes linearly from 10 A to 2 A in t seconds. If the induced emf is 40 V, what is the time t?
What are the materials which exhibit electric polarization even in the absence of an applied electrical field?
What is the relative permittivity of slate dielectric?
- Static electricity is produced by ________.
