Question

It has been given that an infinite line charge is producing a field of $9\times {{10}^{4}}N{{C}^{-1}}$ at a distance of $2cm$. Find out the linear charge density.

Answer 1

Hint: The electric field produced will be equivalent to the ratio of linear charge density to the product of twice the pi, electrical permittivity and the distance between the points. Rearrange this equation in terms of the linear charge density and substitute the values in this equation. This will help you in answering this question.

Complete step by step solution:
Electric field which has been produced by the infinite line charges at a distance $d$ having linear charge density $\lambda $ is mentioned as the relation,
The electric field produced will be equivalent to the ratio of linear charge density to the product of twice the pi, electrical permittivity and the distance between the points. This can be written as an equation given as,
$E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}d}$
Where ${{\varepsilon }_{0}}$ be the permittivity of the free space.
Rearranging this equation can be written as,
$\lambda =2\pi {{\varepsilon }_{0}}dE$
It has been already mentioned in the question that the distance of the charge from the field produced will be,
$d=2cm=0.02m$
The electric field produced here can be shown as,
$E=9\times {{10}^{4}}N{{C}^{-1}}$
As we all know the value the very famous constant,
$\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$
Substituting all these values in these equation can be shown as,
$\lambda =\dfrac{0.02\times 9\times {{10}^{4}}}{2\times 9\times {{10}^{9}}}$
Simplifying this equation can be shown as,
$\lambda =10\mu C{{m}^{-1}}$
Therefore the linear charge density has been calculated.

Note: Linear charge density can be defined as the quantity of charge per unit length, which will be measured in coulombs per meter at any points on a line charge distribution. Charge density will be either positive or negative. This is because the electric charge can have the values to be either negative or positive.