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Q11P

Expert-verifiedFound in: Page 653

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Two charged particles are fixed to an x axis: Particle 1of charge **

The net electric field produced by the two particles is equal to zero at.$-30\text{\hspace{0.17em}cm}$

- Charge of particle 1 at position $x=20\text{cm}$ is${q}_{1}=2.1\times {10}^{-8}\text{\hspace{0.17em}C}$
- Charge of particle 2 at the position$x=70\text{cm}$, is ${q}_{2}=-8.4\times {10}^{-8}\text{\hspace{0.17em}C}$

According to the superposition principle, the net electric field at a point, due to multiple charges present around it, is the vector sum of the fields due to individual charges at that point.Using the superposition law, we can get the value of the net electric field and the required value of the x-coordinate at which this field is zero.

The magnitude of the electric field,

$E=\frac{q}{4\pi {\epsilon}_{o}{R}^{2}}\widehat{R}$ (i)

Where is the distance of the field point from the charge and q is the charge on the particle.

According to the superposition principle, the electric field at a point due to more than one charges,

$\overrightarrow{E}=\underset{i=1}{\overset{n}{}}\overrightarrow{{E}_{i}}=\underset{i=1}{\overset{n}{}}\frac{{q}_{i}}{4\pi {\epsilon}_{o}{{r}_{i}}^{2}}\widehat{{r}_{i}}$ (ii)

Here, ${q}_{i}$ are the multiple charges present, ${r}_{i}$ is the distance between individual charges and the point where field needs to be identified.

Let x be the distance of point P along x-axis, the point where the field vanishes.Then, the net electric field at P using equation (ii) in equation (i) is given by:

$E=\frac{1}{4\pi {\epsilon}_{o}}\left(\frac{\left|{q}_{2}\right|}{{(x-{x}_{2})}^{2}}-\frac{\left|{q}_{1}\right|}{{(x-{x}_{1})}^{2}}\right)$

If the field is to vanish, then the above equation can be equated to zero to get the required position of the coordinates as follows:

$\begin{array}{l}\frac{\left|{q}_{2}\right|}{{(x-{x}_{2})}^{2}}=\frac{\left|{q}_{1}\right|}{{(x-{x}_{1})}^{2}}\\ \frac{\left|{q}_{2}\right|}{\left|{q}_{1}\right|}=\frac{{(x-{x}_{2})}^{2}}{{(x-{x}_{1})}^{2}}\\ \frac{x-70\text{cm}}{x-20\text{cm}}=\pm 2.0\\ x=-30\text{cm}\end{array}$

The results are depicted in the figure below. At P, the field ${\overrightarrow{E}}_{1}$ due to points to${q}_{1}$ the left, while the field ${\overrightarrow{E}}_{2}$due to${q}_{2}$ charge points to the right. Since the magnitude of both${\overrightarrow{E}}_{1}$ and ${\overrightarrow{E}}_{2}$aresame, so net field at P is zero.

Hence, the distance of the point where field is zero, from the origin, is.$-30\text{\hspace{0.17em}cm}$

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