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Found in: Page 625

### Fundamentals Of Physics

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

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# In Fig. 21-26, particle 1 of charge $\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbit{\mu }\mathbit{C}$ and particle 2 of charge $\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{ }\mathbit{\mu }\mathbit{C}\mathbf{}$are held at separation L=10.0cm on an x-axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

1. The x-coordinate of particle 3 for the net force on it to be zero is 14cm
2. The y-coordinate of particle 3 for the net force on it to be zero is 0cm

See the step by step solution

## Step 1: The given data

The values of charges on the three particles are:

${q}_{1}=+1.0\mu C,{q}_{2}=-3.0\mu C, and {q}_{3}=unknown$

The separation between particles 1 and 2, $L=10.0cm$

The net force on the particle from particles 1 and 2 is zero.

## Step 2: Understanding the concept of Coulomb’s law

Using the concept of Coulomb's law, we can get the required equation to solve to get the coordinates of particle 3 for the condition of the net force acting on it as zero.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{|{q}_{1}||{q}_{2}|}{{r}^{2}}$ (1)

## Step 3: a) Calculation of the x-coordinate of the location of particle 3

There is no equilibrium position for q3 between the two fixed charges because it is being pulled by one and pushed by the other (since q1 and q2 have different signs); in this region, this means the two force arrows on q3 are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis that is nearest q2 and furthest from q1 . an equilibrium position for q3 cannot be found because |q1|$<$|q2| and the magnitude of the force exerted by q2 is everywhere (in that region) stronger than that exerted by q1 on q3. Thus, we must look in the semi-infinite region of the axis which is nearest q1 and furthest from q2, where the net force on q3 has magnitude using equation (1) as follows:

${F}_{3}={F}_{13}+{F}_{23}$

$=\left|k\frac{\left|{q}_{1}{q}_{3}\right|}{{L}_{0}^{2}}-k\frac{\left|{q}_{2}{q}_{3}\right|}{{\left(L+{L}_{0}\right)}^{2}}\right|.....................\left(2\right)$

with L = 10 cm and L0 is assumed to be positive. We set this equal to zero, as required by the problem. Now, we obtain the equation (2) by canceling the equal terms as given: $\frac{|{q}_{1}|}{\left({L}_{0}^{2}\right)}-\frac{|{q}_{2}|}{\left(L+{L}_{0}{\right)}^{2}}=0\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{{L}_{0}}\right)}^{2}=|\frac{{q}_{2}}{{q}_{1}}|\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{L0}\right)}^{2}=|\left(-3.0\mu C\right)/\left(-1.0\mu C\right)|\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{{L}_{0}}\right)}^{}=3.0$

Therefore, we get,

localid="1662643008359" $\left(L+{L}_{0}\right)/{L}_{0}=\surd 3\phantom{\rule{0ex}{0ex}}{L}_{0}=\frac{L}{\left(\surd 3-1\right)}\phantom{\rule{0ex}{0ex}}{L}_{0}=\frac{10cm}{\left(\surd 3-1\right)}$

L0=14cm

Hence, the value of the x-coordinate of particle 3 is 14cm

## Step 4: b) Calculation of the y-coordinate of the location of particle 3

As stated above in the calculations of part (a), y = 0

Hence, the value of the y-coordinate of particle 3 is 0cm.

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