two infinite parallel lines of charge are shown in the figure below ,a

1.	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» 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 $-24 \times 10^3 \frac{N}{C}$ .