Question

For an infinitely long metal cylinder, the radius is 3 mm, $K=6.28$ and charge density$=4\mu C/{{m}^{2}}$. What is the electric intensity (E) at a distance of 1.5 m from the axis?
Given:$\left[ \dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}} \right]$
(A) 144 N/C
(B) 160 N/C
(C) 288 N/C
(D)72 N/C

Answer 1

We have a specific formula to deal with such kind of problems. We got all the values required for the answer. We just need to apply the formula to get the desired electric field intensity (E).
Formula used:
$E=\dfrac{\sigma R}{{{\varepsilon }_{0}}Kr}$
Here,
$E=$Required electric field intensity.
$\sigma =$ Charge density.
$R=$Radius of cylinder.
$K=$6.28.
$r\,=$Distance from axis.

Complete step by step answer:
We use the given equation;
$E=\dfrac{\sigma R}{{{\varepsilon }_{0}}Kr}$
Where,
$E=$Required electric field intensity.
$\sigma =4\mu C/{{m}^{2}}=4\times {{10}^{-6}}C/{{m}^{2}}$ Charge density.
$R=3\,mm=3\times {{10}^{-3}}m=$Radius of cylinder.
$K=$6.28.
$r\,=1.5\,m=$Distance from axis.
$E=\dfrac{\sigma R}{{{\varepsilon }_{0}}Kr}$
$E=\dfrac{\sigma R\times 4\pi }{4\pi {{\varepsilon }_{0}}Kr}$
Multiplying the numerator and denominator by$4\pi$.
Here $\pi =3.14$
Putting all values we get:
$E=\dfrac{\sigma R\times 4\pi }{4\pi {{\varepsilon }_{0}}Kr}$
$E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{\sigma R\times 4\pi }{Kr}$
$E=9\times {{10}^{9}}\times \dfrac{4\times {{10}^{-6}}\times 3\times {{10}^{-3}}\times 4\times 3.14}{6.28\times 1.5}$
$E=144\,N/C$
Hence the right answer for this question is option A.

Note: The space around an electric charge in which its influence can be felt is known as the electric field. The electric field Intensity at a point is the force experienced by a unit positive charge placed at that point.
1)Electric Field Intensity is a vector quantity.
2)It is denoted by ‘E’
The electric field of an infinite cylinder of uniform volume charge density can be obtained by using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r > R, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.