Question

Two long thin parallel wires having linear charge density -3 x 10-9C/m and 4 x 10- 9C/m are kept 20 cm in air. Find the magnitude and direction of electric field at a point 10cm from each wire .

Answer 1

Given that,

First linear charge density $\lambda= -3\times10^{-9}\ C/m$

Second linear charge density $\lambda= 4\times10^{-9}\ C/m$

Distance = 20 cm

We need to calculate the electric field

Using formula of electric field

$E_{1}=\dfrac{\lambda}{2\pi r\epsilon_{0}}$

Put the value into the formula

$E_{1}=\dfrac{-3\times10^{9}}{2\pi\times10\times8.85\times10^{-12}}$

$E_{1}=-5.39\times10^{18}\ N/C$

We need to calculate the electric field

Using formula of electric field

$E_{2}=\dfrac{\lambda}{2\pi r\epsilon_{0}}$

Put the value into the formula

$E_{2}=\dfrac{4\times10^{9}}{2\pi\times10\times8.85\times10^{-12}}$

$E_{2}=7.19\times10^{18}\ N/C$

We need to calculate the net electric field

Using formula of net electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=-5.39\times10^{18}+7.19\times10^{18}$

$E=1.8\times10^{18}\ N/C$

Hence, The magnitude of electric field is $1.8\times10^{18}\ N/C$ and direction of electric field is outward.