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Q13P

Expert-verifiedFound in: Page 625

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 21-26, particle 1 of charge $\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu}\mathit{C}$**** and particle 2 of charge $\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathit{\mu}\mathit{C}\mathbf{}$****are held at separation L=10.0cm** ** on an x-axis. If particle 3 of unknown charge q_{3}**

** **

** **

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

** **

The values of charges on the three particles are:

${q}_{1}=+1.0\mu C,{q}_{2}=-3.0\mu C,\hspace{0.33em}and\hspace{0.33em}{q}_{3}=unknown$

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

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

**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,

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

There is no equilibrium position for q_{3} between the two fixed charges because it is being pulled by one and pushed by the other (since q_{1 }and q_{2} have different signs); in this region, this means the two force arrows on q_{3 }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 q_{2 }and furthest from q_{1 }. an equilibrium position for q_{3 }cannot be found because |q_{1}|$<$|q_{2}| and the magnitude of the force exerted by q_{2 }is everywhere (in that region) stronger than that exerted by q_{1 }on q_{3}. Thus, we must look in the semi-infinite region of the axis which is nearest q_{1 }and furthest from q_{2}, where the net force on q_{3 }has magnitude using equation (1) as follows:

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

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

with L = 10 cm and L_{0} 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: $\frac{\left|{q}_{1}\right|}{\left({L}_{0}^{2}\right)}-\frac{\left|{q}_{2}\right|}{(L+{L}_{0}{)}^{2}}=0\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{{L}_{0}}\right)}^{2}=\left|\frac{{q}_{2}}{{q}_{1}}\right|\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{L0}\right)}^{2}=\left|\right(-3.0\mu C)/(-1.0\mu C\left)\right|\phantom{\rule{0ex}{0ex}}{\left(\frac{L+{L}_{0}}{{L}_{0}}\right)}^{}=3.0$

Therefore, we get,

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

L_{0}=14cm_{ }

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

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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