Question

The electric field at a point 2cm from an infinite line charge of linear charge density \[{10^{ - 7}}C.{m^{ - 1}}\] is?
A) \[4.5 \times {10^4}N{C^{ - 1}}\]
B) \[9 \times {10^4}N{C^{ - 1}}\]
C) \[9 \times {10^2}N{C^{ - 1}}\]
D) \[18 \times {10^4}N{C^{ - 1}}\]

Answer 1

Hint:For an infinite line charge with uniform charge density, the electric field at some distance can be calculated using Gauss law. The electric field is always perpendicular to the line charge and directed outward for positive charge density.

Formula used:
Value of electric field at a distance r from an infinite line charge of uniform charge density is given by:
$E = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}}$ ……………..(1)
Where,
E is the electric field value,
$\lambda $ is the linear charge density of the line charge,
r is the distance of the point from the line,
${\varepsilon _0}$ is the permittivity of free space, ${\varepsilon _0} = 8.85 \times {10^{ - 12}}F{m^{ - 1}}$.

Complete step by step answer:
Given:
Linear charge density of the line charge is $\lambda = {10^{ - 7}}C.{m^{ - 1}}$.
Distance of the point from the line charge is $r = 2cm = 0.02m$.

To find: The electric field value at the given point.

Step 1
Substitute the given values of $\lambda $, r and ${\varepsilon _0}$ in eq.(1) to get the value of E as:
$
  E = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}} = \dfrac{{{{10}^{ - 7}}C.{m^{ - 1}}}}{{2\pi \times 8.85 \times {{10}^{ - 12}}F{m^{ - 1}} \times 0.02m}} \\
  \therefore E = 9 \times {10^4}N{C^{ - 1}} \\
 $

As here, the given linear charge density is positive for the infinite long line charge so the direction of the electric field will be directed perpendicularly outward from the line charge.

Correct answer:
The electric field at a point 2cm from an infinite line charge will be (b) \[9 \times {10^4}N{C^{ - 1}}\].

Note: While deriving the given formula in eq.(1) it was assumed that the charge density is uniform and the line charge is infinite. Due to this infinite long assumption we can safely take the electric field to be perpendicularly outward and use that in Gauss law to derive eq.(1). In practice nothing is infinite, hence, near the finite boundaries the electric fields don’t remain exactly perpendicular and this formula can’t be used there anymore.