Question

An infinite line charge produces a field of $9 \times {10^4}N/C$ at a distance of $2$ cm. Calculate the linear charge density.

Answer 1

Hint
As the Electric field produced by the infinite line charges at a distance d having linear charge density $\lambda$ is given by the relation i.e. $E = \dfrac{\lambda }{{2\pi { \in _0}d}}$, on substituting the values we get the required value of the linear charge density.

Complete step by step solution
Now, firstly we will write the given entities from the given problem
Electric field produced is $E = 9 \times {10^4}N/C$
The distance of the point from infinite line charge is $d = 2cm = 0.02m$
As we know the formula for electric field produced by an infinite line charge is
$E = \dfrac{\lambda }{{2\pi { \in _0}d}}$ ……………….. (1)
Where, E is the produced electric field
ϵ0 is the permittivity of free space
d is the distance of the point from the line charge density
we also know that, $\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
so, $\pi { \in _0} = \dfrac{1}{{4 \times 9 \times {{10}^9}}}$
now, use this value in the equation (1), we get
$\Rightarrow \lambda = 2\pi { \in _0}d \times E$
Substitute the given values, we get
$\Rightarrow \lambda = 2 \times \dfrac{1}{{4 \times 9 \times {{10}^9}}} \times 0.02 \times 9 \times {10^4}$
$\Rightarrow \lambda = {10^{ - 7}}C{m^{ - 1}}$
Hence, the line charge density is $\lambda = {10^{ - 7}}C{m^{ - 1}}$.

Note
Here, $ϵ_0$ is the permittivity of free space. Other names of the quantities are electric constant or distributed capacitance of the vacuum. It is an ideal physical constant. It is the value of absolute dielectric permittivity of vacuum.