Question

two infinite parallel lines of charge are shown in the figure below ,and are separated by 5 cm.Their uniform linear charge densities are (λ1) +2.0x10^-6 C/m and (λ2)=-3.0x10^-6 C/m.Find the net electric field at point that is 1.0 cm to the left of (λ1).Express your answer in unit vector notation

Answer 1

Given :

The separation between the two infinite parallel lines of charge =5 cm

The uniform linear charge density of first line is :

$\lambda_1 = +2.0 \times 10^{-6} \frac{C}{m}$

The uniform linear charge density of 2nd line is :

$\lambda_2 =-3.0 \times 10^{-6} \frac{C}{m}$

To Find :

The net electric field at a point 1 cm to the left of $\lambda_1$= ?

Solution :

The net electric field at the point 1 cm left of $\lambda_1$ will be = $E_1 - E_2$

Here $E_1$  is electric field at this point due to $\lambda_1$

Here $E_2$  is electric field at this point due to $\lambda_2$

Since , the electric field due to an infinitely long line charge is =$\frac{k\lambda }{r}$

Here k is electrostatic constant , $\lambda$ is linear charge density and r is distance

of point .

So, $E_{net} = \frac{k\lambda_1}{r_1} -\frac{k\lambda_2}{r_2}$

Or, $E_{net} = k(\frac{\ambda_1}{r_1}- \frac{\lambda_2}{r_2})$

              =$9 \times 10^9(\frac{2\times 10^{-6}}{5+1} -\frac{3\times 10^{-6}}{1})$

              = $9 \times 10^9 \times 10^{-6} (\frac{2}{6}- \frac{3}{1})$

              = $9 \times 10^3 (\frac{1-9}{3})$

              = $3 \times 10^3 \times (-8)$

              = $-24 \times 10^3 \frac{N}{C}$

Hence ,the electric at the required point is towards right and its magnitude is $-24 \times 10^3 \frac{N}{C}$ .