Question

In Fig. 23-43, short sections of two very long parallel lines of charge are shown, fixed in place, separated by L=8.00 cmThe uniform linear charge densities are$\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\mu C}\mathbf{/}\mathbf{m}$for line 1 and$\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu C}\mathbf{/}\mathbf{m}$for line 2. Where along the x-axis shown is the net electric field from the two lines zero?

Short answer

The net electric field from the two lines is zero at 8.00 along the x-axis.

Step-by-step solution

The given data

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

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

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

Understanding the concept of the electric field

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,$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ (1)

Calculation of the position on the x-axis for the net field to be zero

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 “line 1” since its magnitude of charge (per unit length) exceeds that of “line 2”; thus, we look in the region to the right of “line 2” for P. Thus, the expression of the net electric field using equation (1) is given as:

$E_{net}=E_1+E_2$

$=\frac{2\lambda }{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{x}+\mathrm{L}/2\right)}+\frac{2{\lambda }_2}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{x}-\mathrm{L}/2\right)}$ …………………(2)

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

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

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