Question

A charge of $5 \times {10^{ - 10}}C$ is given to a metal cylinder of length 10m, placed in air. The electric intensity at a distance of 0.2m from its axis is:
A. 4.5V/m
B. 45V/m
C. 450V/m
D. 100V/m

Answer 1

Hint: The electric field intensity is the force experienced by a unit positive charge which is placed at a specific point. It is also known as the magnitude of the electric field at that point. It can be calculated using the below formula which involves linear charge density. Use that formula to find the electric intensity.

Formula Used: Electric field intensity E is $E = \dfrac{\lambda }{{2\pi r{\varepsilon _0}}}$, where $\lambda $ is the linear charge density, ${\varepsilon _0}$ is the electric constant, r is the distance from the axis.
Linear charge density $\lambda $ is $\lambda = \dfrac{Q}{l}$, where Q is the charge of the body and l is its length.

Complete step by step answer:We are given that a charge of $5 \times {10^{ - 10}}C$ is given to a metal cylinder of length 10m, placed in air.
We have to find the electric field intensity at a point whose distance is 0.2m from the axis.
Let the point be X whose distance is 0.2m from the cylinder’s axis.

Electric field intensity can be calculated using the formula $E = \dfrac{\lambda }{{2\pi r{\varepsilon _0}}}$
So to calculate electric intensity first we have to get the value of linear charge density.
Linear charge density $\lambda $ is calculated using the formula $\lambda = \dfrac{Q}{l}$
$
  Q = 5 \times {10^{ - 10}}C,l = 10m \\
   \Rightarrow \lambda = \dfrac{{5 \times {{10}^{ - 10}}}}{{10}} \\
  \therefore \lambda = 5 \times {10^{ - 11}}C/m \\
 $
On substituting the values of $\lambda $, r, ${\varepsilon _0}$ in $E = \dfrac{\lambda }{{2\pi r{\varepsilon _0}}}$, we get
$
  {\varepsilon _0} = 8.85 \times {10^{ - 12}},\lambda = 5 \times {10^{ - 11}},\pi = 3.14,r = 0.2 \\
   \Rightarrow E = \dfrac{{5 \times {{10}^{ - 11}}}}{{2 \times 3.14 \times 0.2 \times 8.85 \times {{10}^{ - 12}}}} \\
   \Rightarrow E = \dfrac{{5 \times 10}}{{11.1156}} \\
  \therefore E = \dfrac{{50}}{{11.1156}} = 4.498 \approx 4.5V/m \\
 $
Hence, the correct option is Option A, 4.5V/m.

Note:Do not confuse Electric field intensity with electric field. The basic difference between them is that an electric field is an area around a charge in which an electrostatic force is exerted by it on other charges whereas the electric field intensity is the strength of electric field at any point in space.