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Found in: Page 653

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

In Fig. 22-41, particle 1 of charge q1 = -5.00q and particle 2 of charge q2=+2.00q are fixed to an x axis. (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.

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.
See the step by step solution

Step 1: 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.

Step 2: 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, $\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^{2}}\stackrel{^}{\mathrm{R}}$ (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" $\stackrel{\to }{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\stackrel{\to }{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^{2}}\stackrel{^}{{\mathrm{r}}_{\mathrm{i}}}$ (ii)

Step 3: 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:

$\stackrel{\to }{{\mathrm{E}}_{\mathrm{net}}}=\left(\left|{\mathrm{E}}_{2}\right|-\left|{\mathrm{E}}_{1}\right|\right)\stackrel{^}{\mathrm{i}}$

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:

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

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

Step 4: b) Sketching the electric field lines between and around the particle

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