StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

14P

Expert-verifiedFound in: Page 653

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 22-41, particle 1 of charge q _{1} = -5.00q**

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

- Charge of particle 1, q
_{1}= - 5.00q. - Charge of particle 2, q
_{2}=+2.00 q - They are fixed on the x-axis.

**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" $\overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{{\mathrm{E}}_{\mathrm{i}}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon}}_{\mathrm{o}}{\mathrm{r}}_{\mathrm{i}}^{2}}\hat{{\mathrm{r}}_{\mathrm{i}}}$ (ii)

In the region, x<0, the fields are opposite but unequal (since E_{1} is everywhere bigger than E_{2} 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 q_{2} as compared to q_{1}. 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

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

94% of StudySmarter users get better grades.

Sign up for free