Question

In Fig. 22-41, particle 1 of charge q₁ = -5.00q and particle 2 of charge q₂=+2.00qare fixed to an xaxis. (a) As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? (b)Sketch the net electric field lines between and around the particles.

Short answer

1. The coordinate at which the net electric field is zero is 2.72 L
2. The net electric field between and around the particle is sketched.

Step-by-step solution

The given data

• Charge of particle 1, q₁= - 5.00q.
• Charge of particle 2, q₂=+2.00 q
• They are fixed on the x-axis.

Understanding the concept of electric field 

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again using the superposition law, we can get the value of the electric field in its direction and this determines the net electric field at that point. Equating the net field to zero will give to the required coordinate.

Formulae:

The magnitude of the electric field,$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$ (i)

where, R = The distance of field point from the charge

q = charge of the particle

According to the superposition principle, the electric field at a point due to more than one charges, data-custom-editor="chemistry" $\overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^2}\widehat{{\mathrm{r}}_{\mathrm{i}}}$ (ii)

Calculation of the coordinate at which the net electric field is zero

In the region, x<0, the fields are opposite but unequal (since E₁ is everywhere bigger than E₂ in this region), so there is no possibility of cancellation of the fields. In the region between the charges 0 < x < L both the fields will add up as they point in the same direction, so net field cannot be zero. In the region x > L, the fields are opposite and unequal, so the net field in this range using equation (ii) is given as:

$\overrightarrow{{\mathrm{E}}_{\mathrm{net}}}=\left(\left|{\mathrm{E}}_2\right|-\left|{\mathrm{E}}_1\right|\right)\hat{\mathrm{i}}$

Though$\left|{\mathrm{q}}_1\right|>\left|{\mathrm{q}}_2\right|$,the net electric field$\left({\mathrm{E}}_{\mathrm{net}}\right)$ can be zero,since these points are closer to q₂ as compared to q₁. Thus, we look for the zero net field point in the region x > L using equation (i) as:

$\begin{array}{rcl}\left|\overrightarrow{{\mathrm{E}}_2}\right|& =& \left|\overrightarrow{{\mathrm{E}}_1}\right|\\ \frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\left|{\mathrm{q}}_1\right|}{{\mathrm{x}}^2}& =& \frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{{\mathrm{q}}_2}{{\left(\mathrm{x}-\mathrm{L}\right)}^2}\\ \mathrm{x}& =& \frac{\mathrm{L}}{1-\sqrt{2/5}}\\ \mathrm{x}& \approx & 2.72\mathrm{L}\end{array}$

Hence, the value of the x-coordinate is 2.72 L

b) Sketching the electric field lines between and around the particle

the field lines around the particles are shown in the figure below: