91影视

a) The separation between the two parallel lines,L=8.00 cm

b) Uniform charge density for line 1,${\mathrm{位}}_1=6脳{10}^{-6}\mathrm{C}/\mathrm{m}$

c) Uniform charge density for line 2,${\mathrm{位}}_2=-2脳{10}^{-6}\mathrm{C}/\mathrm{m}$

Using the concept of the electric field, we can get the individual electric fields of the two long parallel lines. Thus, for the net electric field to be zero; we get the expression for x. Now, solving it, we can get the required value of x.

Formula:

The electric field of a long line,$\mathrm{E}=\frac{\mathrm{位}}{2{\mathrm{蟺蔚}}_0\mathrm{r}}$ (1)

We reason that point P (the point on the x-axis where the net electric field is zero) cannot be between the lines of charge (since their charges have an opposite sign). We reason further that P is not to the left of 鈥渓ine 1鈥� since its magnitude of charge (per unit length) exceeds that of 鈥渓ine 2鈥�; thus, we look in the region to the right of 鈥渓ine 2鈥� for P. Thus, the expression of the net electric field using equation (1) is given as:

$E_{net}=E_1+E_2$

$=\frac{2位}{4{\mathrm{蟺蔚}}_0\left(\mathrm{x}+\mathrm{L}/2\right)}+\frac{2{位}_2}{4{\mathrm{蟺蔚}}_0\left(\mathrm{x}-\mathrm{L}/2\right)}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌�(2)

Setting this equal to zero and solving for x we find the value as follows:

$$\begin{gathered} \mathrm{x}=\left(\frac{{\mathrm{位}}_1-{\mathrm{位}}_2}{{\mathrm{位}}_1-{\mathrm{位}}_2}\right)\left(\mathrm{L}/2\right) \\ =\left(\frac{6.0\mathrm{渭颁}/\mathrm{m}-\left(2.0\mathrm{C}/\mathrm{m}\right)}{6.0\mathrm{渭颁}/\mathrm{m}+\left(-2.0\mathrm{C}/\mathrm{m}\right)}\right)\frac{8.0\mathrm{cm}}{2} \\ =8.0\mathrm{cm} \end{gathered}$$

Hence, the value of the position of the point on the x-axis is 8.0 cm .